Structure of Atom — NCERT Solutions

(i) Given: Mass of one electron = 9.10939 × 10⁻²⁸ g

Number of electrons that weigh 1 g:
$$N = \frac{1\text{ g}}{9.10939 \times 10^{-28}\text{ g}} = 1.098 \times 10^{27} \text{ electrons}$$

(ii) Mass of one mole of electrons:
$$m = 6.022 \times 10^{23} \times 9.10939 \times 10^{-31}\text{ kg} = 5.486 \times 10^{-7}\text{ kg} \approx 5.49 \times 10^{-4}\text{ g}$$

Charge of one mole of electrons:
$$Q = 6.022 \times 10^{23} \times 1.6022 \times 10^{-19}\text{ C} = 9.65 \times 10^{4}\text{ C} = 96500\text{ C (1 Faraday)}$$

(i) Methane (CH₄): electrons per molecule = 6 (C) + 4×1 (H) = 10 electrons

Total electrons in 1 mole of CH₄:
$$N = 10 \times 6.022 \times 10^{23} = 6.022 \times 10^{24}\text{ electrons}$$

(ii) ¹⁴C has 8 neutrons per atom. Molar mass of ¹⁴C = 14 g/mol.

Moles of ¹⁴C in 7 mg:
$$n = \frac{7 \times 10^{-3}\text{ g}}{14\text{ g/mol}} = 5 \times 10^{-4}\text{ mol}$$

(a) Total number of neutrons:
$$= 5 \times 10^{-4} \times 6.022 \times 10^{23} \times 8 = 2.409 \times 10^{21}\text{ neutrons}$$

(b) Total mass of neutrons:
$$= 2.409 \times 10^{21} \times 1.675 \times 10^{-27}\text{ kg} = 4.035 \times 10^{-6}\text{ kg}$$

(iii) NH₃ has molar mass = 17 g/mol. Protons per molecule = 7 (N) + 3×1 (H) = 10 protons.

Moles of NH₃ in 34 mg:
$$n = \frac{34 \times 10^{-3}\text{ g}}{17\text{ g/mol}} = 2 \times 10^{-3}\text{ mol}$$

(a) Total number of protons:
$$= 2 \times 10^{-3} \times 6.022 \times 10^{23} \times 10 = 1.2044 \times 10^{22}\text{ protons}$$

(b) Total mass of protons:
$$= 1.2044 \times 10^{22} \times 1.6726 \times 10^{-27}\text{ kg} = 2.014 \times 10^{-5}\text{ kg}$$

The answer will NOT change with temperature and pressure because the number of moles (and hence number of protons) depends only on the mass of the sample, not on T and P.

For any nucleus $^A_Z X$: Number of protons = Z, Number of neutrons = A − Z.

$$^{13}_6\text{C}: \text{Protons} = 6,\quad \text{Neutrons} = 13 - 6 = 7$$

$$^{16}_8\text{O}: \text{Protons} = 8,\quad \text{Neutrons} = 16 - 8 = 8$$

$$^{24}_{12}\text{Mg}: \text{Protons} = 12,\quad \text{Neutrons} = 24 - 12 = 12$$

$$^{56}_{26}\text{Fe}: \text{Protons} = 26,\quad \text{Neutrons} = 56 - 26 = 30$$

$$^{88}_{38}\text{Sr}: \text{Protons} = 38,\quad \text{Neutrons} = 88 - 38 = 50$$

The complete symbol is $^A_Z X$ where X is the element symbol.

(i) Z = 17 → Chlorine (Cl); A = 35:
$$^{35}_{17}\text{Cl}$$

(ii) Z = 92 → Uranium (U); A = 233:
$$^{233}_{92}\text{U}$$

(iii) Z = 4 → Beryllium (Be); A = 9:
$$^{9}_{4}\text{Be}$$

Given: $\lambda = 580\text{ nm} = 580 \times 10^{-9}\text{ m}$, $c = 3 \times 10^8\text{ m s}^{-1}$

Frequency:
$$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8\text{ m s}^{-1}}{580 \times 10^{-9}\text{ m}} = 5.17 \times 10^{14}\text{ Hz}$$

Wavenumber:
$$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{580 \times 10^{-9}\text{ m}} = 1.724 \times 10^{6}\text{ m}^{-1} = 1.724 \times 10^{4}\text{ cm}^{-1}$$

Using $E = h\nu$ and $E = hc/\lambda$; $h = 6.626 \times 10^{-34}\text{ J s}$

(i) $\nu = 3 \times 10^{15}\text{ Hz}$:
$$E = h\nu = 6.626 \times 10^{-34} \times 3 \times 10^{15} = 1.988 \times 10^{-18}\text{ J}$$

(ii) $\lambda = 0.50\text{ Å} = 0.50 \times 10^{-10}\text{ m} = 5 \times 10^{-11}\text{ m}$:
$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{5 \times 10^{-11}} = 3.976 \times 10^{-15}\text{ J} \approx 3.98 \times 10^{-15}\text{ J}$$

Given: Time period $T = 2.0 \times 10^{-10}\text{ s}$

Frequency:
$$\nu = \frac{1}{T} = \frac{1}{2.0 \times 10^{-10}} = 5.0 \times 10^{9}\text{ Hz}$$

Wavelength:
$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{5.0 \times 10^9} = 6.0 \times 10^{-2}\text{ m} = 0.06\text{ m}$$

Wavenumber:
$$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{6.0 \times 10^{-2}} = 16.67\text{ m}^{-1} = 0.1667\text{ cm}^{-1}$$

Given: $\lambda = 4000\text{ pm} = 4000 \times 10^{-12}\text{ m} = 4 \times 10^{-9}\text{ m}$

Energy of one photon:
$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-9}} = 4.969 \times 10^{-17}\text{ J}$$

Number of photons:
$$n = \frac{1\text{ J}}{4.969 \times 10^{-17}\text{ J}} = 2.012 \times 10^{16} \approx 2.0 \times 10^{16}\text{ photons}$$

Given: $\lambda = 4 \times 10^{-7}\text{ m}$, $W_0 = 2.13\text{ eV}$

(i) Energy of photon:
$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} = 4.969 \times 10^{-19}\text{ J}$$
$$E = \frac{4.969 \times 10^{-19}}{1.6020 \times 10^{-19}} = 3.10\text{ eV}$$

(ii) Kinetic energy of emitted electron:
$$KE = E - W_0 = 3.10 - 2.13 = 0.97\text{ eV}$$
$$KE = 0.97 \times 1.6020 \times 10^{-19} = 1.554 \times 10^{-19}\text{ J}$$

(iii) Velocity of photoelectron ($m_e = 9.109 \times 10^{-31}\text{ kg}$):
$$KE = \frac{1}{2}m_e v^2 \Rightarrow v = \sqrt{\frac{2 \times KE}{m_e}}$$
$$v = \sqrt{\frac{2 \times 1.554 \times 10^{-19}}{9.109 \times 10^{-31}}} = \sqrt{3.413 \times 10^{11}} = 5.84 \times 10^{5}\text{ m s}^{-1}$$

Given: $\lambda = 242\text{ nm} = 242 \times 10^{-9}\text{ m}$

Energy per photon:
$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{242 \times 10^{-9}} = 8.21 \times 10^{-19}\text{ J}$$

Ionisation energy per mole:
$$IE = E \times N_A = 8.21 \times 10^{-19} \times 6.022 \times 10^{23} = 4.945 \times 10^{5}\text{ J mol}^{-1}$$
$$\boxed{IE = 494.5\text{ kJ mol}^{-1}}$$

Given: Power $P = 25\text{ W} = 25\text{ J s}^{-1}$, $\lambda = 0.57\text{ μm} = 0.57 \times 10^{-6}\text{ m}$

Energy per photon:
$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.57 \times 10^{-6}} = 3.487 \times 10^{-19}\text{ J}$$

Rate of emission:
$$n = \frac{P}{E} = \frac{25}{3.487 \times 10^{-19}} = 7.169 \times 10^{19} \approx 7.17 \times 10^{19}\text{ quanta s}^{-1}$$

Given: $\lambda_0 = 6800\text{ Å} = 6800 \times 10^{-10}\text{ m} = 6.8 \times 10^{-7}\text{ m}$

Since electrons are emitted with zero velocity, the incident wavelength equals the threshold wavelength.

Threshold frequency:
$$\nu_0 = \frac{c}{\lambda_0} = \frac{3 \times 10^8}{6.8 \times 10^{-7}} = 4.41 \times 10^{14}\text{ Hz}$$

Work function:
$$W_0 = h\nu_0 = 6.626 \times 10^{-34} \times 4.41 \times 10^{14} = 2.922 \times 10^{-19}\text{ J}$$

Using the Rydberg formula:
$$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

where $R_H = 1.097 \times 10^7\text{ m}^{-1}$, $n_1 = 2$, $n_2 = 4$:

$$\frac{1}{\lambda} = 1.097 \times 10^7\left(\frac{1}{4} - \frac{1}{16}\right) = 1.097 \times 10^7 \times \frac{3}{16}$$

$$\frac{1}{\lambda} = 1.097 \times 10^7 \times 0.1875 = 2.057 \times 10^6\text{ m}^{-1}$$

$$\lambda = \frac{1}{2.057 \times 10^6} = 4.86 \times 10^{-7}\text{ m} = 486\text{ nm}$$

This is the blue-green line in the Balmer series (visible region).

Energy of electron in nth orbit of H atom:
$$E_n = \frac{-2.18 \times 10^{-18}}{n^2}\text{ J}$$

For n = 5:
$$E_5 = \frac{-2.18 \times 10^{-18}}{25} = -8.72 \times 10^{-20}\text{ J}$$

Energy required to ionise from n = 5 (remove to n = ∞, where E = 0):
$$\Delta E = 0 - (-8.72 \times 10^{-20}) = 8.72 \times 10^{-20}\text{ J}$$

For n = 1:
$$E_1 = -2.18 \times 10^{-18}\text{ J}$$
$$\Delta E_1 = 2.18 \times 10^{-18}\text{ J}$$

Comparison:
$$\frac{\Delta E_1}{\Delta E_5} = \frac{2.18 \times 10^{-18}}{8.72 \times 10^{-20}} = 25$$

The energy required to ionise from n = 1 is 25 times greater than from n = 5.

When an electron drops from the nth level to the ground state, the maximum number of spectral lines is given by:
$$\text{Number of lines} = \frac{n(n-1)}{2}$$

For n = 6:
$$\text{Number of lines} = \frac{6(6-1)}{2} = \frac{6 \times 5}{2} = 15$$

The 15 possible transitions are: 6→5, 6→4, 6→3, 6→2, 6→1, 5→4, 5→3, 5→2, 5→1, 4→3, 4→2, 4→1, 3→2, 3→1, 2→1.

(i) Energy of nth orbit: $E_n = E_1/n^2$

$$E_5 = \frac{-2.18 \times 10^{-18}}{5^2} = \frac{-2.18 \times 10^{-18}}{25} = -8.72 \times 10^{-20}\text{ J atom}^{-1}$$

(ii) Radius of nth Bohr orbit for hydrogen: $r_n = 0.0529 \times n^2\text{ nm}$

$$r_5 = 0.0529 \times 5^2 = 0.0529 \times 25 = 1.3225\text{ nm} = 132.25\text{ pm}$$

The Balmer series has $n_1 = 2$. Longest wavelength corresponds to smallest energy transition, i.e., $n_2 = 3$.

$$\bar{\nu} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = 1.097 \times 10^7\left(\frac{1}{4} - \frac{1}{9}\right)$$

$$= 1.097 \times 10^7 \times \frac{9-4}{36} = 1.097 \times 10^7 \times \frac{5}{36}$$

$$= 1.097 \times 10^7 \times 0.1389 = 1.523 \times 10^6\text{ m}^{-1}$$

$$\bar{\nu} = 1.523 \times 10^4\text{ cm}^{-1}$$

Given: $E_1 = -2.18 \times 10^{-11}\text{ ergs} = -2.18 \times 10^{-18}\text{ J}$

$$E_5 = \frac{E_1}{25} = \frac{-2.18 \times 10^{-18}}{25} = -8.72 \times 10^{-20}\text{ J}$$

Energy required to shift from n=1 to n=5:
$$\Delta E = E_5 - E_1 = -8.72 \times 10^{-20} - (-2.18 \times 10^{-18})$$
$$= 2.18 \times 10^{-18} - 8.72 \times 10^{-20} = 2.0928 \times 10^{-18}\text{ J}$$

Wavelength of light emitted when electron returns to ground state (n=5 → n=1):
$$\lambda = \frac{hc}{\Delta E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{2.0928 \times 10^{-18}}$$
$$= \frac{1.988 \times 10^{-25}}{2.0928 \times 10^{-18}} = 9.498 \times 10^{-8}\text{ m} \approx 9.5 \times 10^{-8}\text{ m} = 95\text{ nm}$$

Energy at n = 2:
$$E_2 = \frac{-2.18 \times 10^{-18}}{4} = -5.45 \times 10^{-19}\text{ J}$$

Energy at n = ∞: $E_\infty = 0$

Energy required:
$$\Delta E = 0 - (-5.45 \times 10^{-19}) = 5.45 \times 10^{-19}\text{ J}$$

Longest wavelength (minimum energy = exactly $\Delta E$):
$$\lambda = \frac{hc}{\Delta E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{5.45 \times 10^{-19}}$$
$$= \frac{1.988 \times 10^{-25}}{5.45 \times 10^{-19}} = 3.647 \times 10^{-7}\text{ m} = 3.647 \times 10^{-5}\text{ cm}$$

Using de Broglie equation: $\lambda = \dfrac{h}{mv}$

Given: $m = 9.109 \times 10^{-31}\text{ kg}$, $v = 2.05 \times 10^7\text{ m s}^{-1}$, $h = 6.626 \times 10^{-34}\text{ J s}$

$$\lambda = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 2.05 \times 10^7}$$

$$= \frac{6.626 \times 10^{-34}}{1.867 \times 10^{-23}} = 3.548 \times 10^{-11}\text{ m} \approx 3.55 \times 10^{-11}\text{ m}$$

Given: $m = 9.1 \times 10^{-31}\text{ kg}$, $KE = 3.0 \times 10^{-25}\text{ J}$

First find velocity:
$$KE = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 3.0 \times 10^{-25}}{9.1 \times 10^{-31}}}$$
$$= \sqrt{6.593 \times 10^5} = 8.12 \times 10^2\text{ m s}^{-1}$$

de Broglie wavelength:
$$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 8.12 \times 10^2}$$
$$= \frac{6.626 \times 10^{-34}}{7.389 \times 10^{-28}} = 8.967 \times 10^{-7}\text{ m} \approx 8.97 \times 10^{-7}\text{ m}$$

Count electrons for each species:

• $\text{Na}^+$: Na has 11 electrons, loses 1 → 10 electrons
• $\text{K}^+$: K has 19 electrons, loses 1 → 18 electrons
• $\text{Mg}^{2+}$: Mg has 12 electrons, loses 2 → 10 electrons
• $\text{Ca}^{2+}$: Ca has 20 electrons, loses 2 → 18 electrons
• $\text{S}^{2-}$: S has 16 electrons, gains 2 → 18 electrons
• $\text{Ar}$: Ar has 18 electrons

Isoelectronic pairs:

• $\text{Na}^+$ and $\text{Mg}^{2+}$ (10 electrons each)
• $\text{K}^+$, $\text{Ca}^{2+}$, $\text{S}^{2-}$, and $\text{Ar}$ (18 electrons each)

(i) Electronic configurations:

(a) $\text{H}^-$: H has 1 electron, gains 1 → 2 electrons
$$\text{H}^-: 1s^2$$

(b) $\text{Na}^+$: Na has 11 electrons, loses 1 → 10 electrons
$$\text{Na}^+: 1s^2\, 2s^2\, 2p^6$$

(c) $\text{O}^{2-}$: O has 8 electrons, gains 2 → 10 electrons
$$\text{O}^{2-}: 1s^2\, 2s^2\, 2p^6$$

(d) $\text{F}^-$: F has 9 electrons, gains 1 → 10 electrons
$$\text{F}^-: 1s^2\, 2s^2\, 2p^6$$

(ii) Atomic numbers:

(a) $3s^1$: Configuration is $1s^2\, 2s^2\, 2p^6\, 3s^1$ → Total electrons = 11 → Z = 11 (Na)

(b) $2p^3$: Configuration is $1s^2\, 2s^2\, 2p^3$ → Total electrons = 7 → Z = 7 (N)

(c) $3p^5$: Configuration is $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^5$ → Total electrons = 17 → Z = 17 (Cl)

(iii) Atoms:

(a) $[\text{He}]\, 2s^1$: $1s^2\, 2s^1$ → 3 electrons → Lithium (Li), Z = 3

(b) $[\text{Ne}]\, 3s^2\, 3p^3$: $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^3$ → 15 electrons → Phosphorus (P), Z = 15

(c) $[\text{Ar}]\, 4s^2\, 3d^1$: $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^1$ → 21 electrons → Scandium (Sc), Z = 21

For g orbitals, $l = 4$ (since s→l=0, p→l=1, d→l=2, f→l=3, g→l=4).

The condition is $l \leq n-1$, so:
$$n - 1 \geq 4 \Rightarrow n \geq 5$$

The lowest value of $n$ that allows g orbitals to exist is $\boxed{n = 5}$.

For a 3d orbital:

• Principal quantum number: $n = 3$
• Azimuthal quantum number: $l = 2$ (d orbital)
• Magnetic quantum number: $m_l = -2, -1, 0, +1, +2$

So the possible values are:
$$n = 3,\quad l = 2,\quad m_l \in \{-2, -1, 0, +1, +2\}$$

(i) In a neutral atom, number of protons = number of electrons.
$$\text{Number of protons} = 29$$

The element is Copper (Cu), Z = 29.

(ii) Electronic configuration of Cu (Z = 29):

Expected: $[\text{Ar}]\, 4s^2\, 3d^9$

Actual (due to extra stability of completely filled d-subshell):
$$\text{Cu}: 1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 3d^{10}\, 4s^1$$
or $[\text{Ar}]\, 3d^{10}\, 4s^1$

• $\text{H}_2^+$: H₂ has 2 electrons, loses 1 → 1 electron
• $\text{H}_2$: Each H has 1 electron → 2 electrons
• $\text{O}_2^+$: O₂ has 16 electrons (2×8), loses 1 → 15 electrons

(i) For n = 3, $l$ can be 0, 1, 2.

• $l = 0$: $m_l = 0$
• $l = 1$: $m_l = -1, 0, +1$
• $l = 2$: $m_l = -2, -1, 0, +1, +2$

(ii) For 3d orbital: $n = 3$, $l = 2$
$$m_l = -2, -1, 0, +1, +2$$

(iii) Checking possibility (condition: $l \leq n-1$):

• 1p: $n=1$, $l=1$; requires $l \leq 0$ → Not possible
• 2s: $n=2$, $l=0$; $0 \leq 1$ → Possible
• 2p: $n=2$, $l=1$; $1 \leq 1$ → Possible
• 3f: $n=3$, $l=3$; requires $l \leq 2$ → Not possible

Notation: $l = 0 \to s$, $l = 1 \to p$, $l = 2 \to d$, $l = 3 \to f$

(a) $n = 1$, $l = 0$ → 1s

(b) $n = 3$, $l = 1$ → 3p

(c) $n = 4$, $l = 2$ → 4d

(d) $n = 4$, $l = 3$ → 4f

(a) $n = 0$, $l = 0$, $m_l = 0$, $m_s = +\frac{1}{2}$
Not possible. The principal quantum number $n$ must be a positive integer ($n \geq 1$). $n = 0$ is not allowed.

(b) $n = 1$, $l = 0$, $m_l = 0$, $m_s = -\frac{1}{2}$
Possible. All values are valid: $l = 0 \leq n-1 = 0$, $m_l = 0$ is within $[-l, +l]$, and $m_s = -\frac{1}{2}$ is valid.

(c) $n = 1$, $l = 1$, $m_l = 0$, $m_s = +\frac{1}{2}$
Not possible. For $n = 1$, $l$ can only be 0 (since $l \leq n-1 = 0$). $l = 1$ is not allowed.

(d) $n = 2$, $l = 1$, $m_l = 0$, $m_s = -\frac{1}{2}$
Possible. $l = 1 \leq n-1 = 1$, $m_l = 0 \in [-1, +1]$, $m_s = -\frac{1}{2}$ is valid.

(e) $n = 3$, $l = 3$, $m_l = -3$, $m_s = +\frac{1}{2}$
Not possible. For $n = 3$, $l$ can be 0, 1, or 2 (since $l \leq n-1 = 2$). $l = 3$ is not allowed.

(f) $n = 3$, $l = 1$, $m_l = 0$, $m_s = +\frac{1}{2}$
Possible. $l = 1 \leq n-1 = 2$, $m_l = 0 \in [-1, +1]$, $m_s = +\frac{1}{2}$ is valid.

(a) $n = 4$, $m_s = -\frac{1}{2}$:

For $n = 4$, total orbitals = $n^2 = 16$. Each orbital can hold one electron with $m_s = -\frac{1}{2}$.
$$\text{Number of electrons} = 16$$

(b) $n = 3$, $l = 0$ (i.e., 3s orbital):

For $l = 0$, $m_l = 0$ only → 1 orbital → can hold 2 electrons.
$$\text{Number of electrons} = 2$$

According to Bohr's second postulate, the angular momentum of an electron in the nth orbit is quantized:
$$m_e v r = \frac{nh}{2\pi} \quad \text{...(1)}$$

The de Broglie wavelength of the electron:
$$\lambda = \frac{h}{m_e v} \Rightarrow m_e v = \frac{h}{\lambda} \quad \text{...(2)}$$

Substituting (2) into (1):
$$\frac{h}{\lambda} \cdot r = \frac{nh}{2\pi}$$

$$r = \frac{n\lambda}{2\pi}$$

$$2\pi r = n\lambda$$

Since $2\pi r$ is the circumference of the orbit:
$$\boxed{\text{Circumference} = n\lambda}$$

This shows that the circumference of the Bohr orbit is an integral multiple ($n$) of the de Broglie wavelength.

For He⁺ (Z = 2), the wavenumber for n = 4 → n = 2 transition:
$$\bar{\nu}_{\text{He}^+} = R_H Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = R_H \times 4 \times \left(\frac{1}{4} - \frac{1}{16}\right) = R_H \times 4 \times \frac{3}{16} = \frac{3R_H}{4}$$

For hydrogen (Z = 1), we need the same wavenumber:
$$\bar{\nu}_{\text{H}} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = \frac{3R_H}{4}$$

$$\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4} = \frac{1}{1} - \frac{1}{4} = \frac{1}{1^2} - \frac{1}{2^2}$$

So $n_1 = 1$ and $n_2 = 2$.

The transition in hydrogen spectrum is n = 2 to n = 1 (Lyman series).

He⁺ is a hydrogen-like species with Z = 2. The energy of the electron in the nth orbit of a hydrogen-like ion is:
$$E_n = -\frac{Z^2 \times 2.18 \times 10^{-18}}{n^2}\text{ J}$$

For He⁺ in ground state (n = 1, Z = 2):
$$E_1(\text{He}^+) = -\frac{4 \times 2.18 \times 10^{-18}}{1} = -8.72 \times 10^{-18}\text{ J}$$

Energy required to remove the electron (ionization):
$$\Delta E = 0 - (-8.72 \times 10^{-18}) = 8.72 \times 10^{-18}\text{ J atom}^{-1}$$