Atoms — NCERT Solutions

For hydrogen atom in a Bohr orbit,
$$E = K+U$$
and from the chapter,
$$K = -E \quad \text{and} \quad U = 2E$$
for a bound Coulomb orbit, with $E$ negative.

Given $E = -13.6\,\text{eV}$:
$$K = -E = 13.6\,\text{eV}$$
$$U = 2E = 2(-13.6) = -27.2\,\text{eV}$$
So the electron's kinetic energy is $13.6\,\text{eV}$ and potential energy is $-27.2\,\text{eV}$.

For hydrogen,
$$E_n = -\frac{13.6}{n^2}\,\text{eV}$$

Ground state: $E_1 = -13.6\,\text{eV}$

For $n=4$:
$$E_4 = -\frac{13.6}{16} = -0.85\,\text{eV}$$

Energy absorbed:
$$\Delta E = E_4 - E_1 = (-0.85)-(-13.6)=12.75\,\text{eV}$$

Convert to joules:
$$\Delta E = 12.75\times 1.6\times 10^{-19} = 2.04\times 10^{-18}\,\text{J}$$

Frequency:
$$\nu = \frac{\Delta E}{h} = \frac{2.04\times 10^{-18}}{6.6\times 10^{-34}} \approx 3.1\times 10^{15}\,\text{Hz}$$

Wavelength:
$$\lambda = \frac{c}{\nu} = \frac{3.0\times 10^8}{3.1\times 10^{15}} \approx 9.7\times 10^{-8}\,\text{m}$$
$$\lambda \approx 97\,\text{nm}$$
So the photon has wavelength $97\,\text{nm}$ and frequency $3.1\times 10^{15}\,\text{Hz}$.

From Bohr's model, the speed in the $n^{\text{th}}$ orbit varies as $1/n$.

For hydrogen:
$$v_n = \frac{2.2\times 10^6}{n}\,\text{m/s}$$

So,

• For $n=1$:

$$v_1 = 2.2\times 10^6\,\text{m/s}$$

• For $n=2$:

$$v_2 = \frac{2.2\times 10^6}{2} = 1.1\times 10^6\,\text{m/s}$$

• For $n=3$:

$$v_3 = \frac{2.2\times 10^6}{3} \approx 7.3\times 10^5\,\text{m/s}$$

The orbital period is
$$T = \frac{2\pi r}{v}$$
For Bohr orbits, $r_n = n^2 a_0$ and $v_n = 2.2\times 10^6/n$ m/s, with $a_0 = 5.3\times 10^{-11}$ m.

For $n=1$

$$T_1 = \frac{2\pi(5.3\times 10^{-11})}{2.2\times 10^6} \approx 1.5\times 10^{-16}\,\text{s}$$

For $n=2$

$$r_2 = 4a_0 = 2.12\times 10^{-10}\,\text{m},\quad v_2 = 1.1\times 10^6\,\text{m/s}$$
$$T_2 = \frac{2\pi(2.12\times 10^{-10})}{1.1\times 10^6} \approx 6.0\times 10^{-16}\,\text{s}$$

For $n=3$

$$r_3 = 9a_0 = 4.77\times 10^{-10}\,\text{m},\quad v_3 = 7.3\times 10^5\,\text{m/s}$$
$$T_3 = \frac{2\pi(4.77\times 10^{-10})}{7.3\times 10^5} \approx 1.4\times 10^{-15}\,\text{s}$$

For hydrogen Bohr orbits,
$$r_n = n^2 r_1$$
where $r_1 = 5.3\times 10^{-11}\,\text{m}$.

For $n=2$:
$$r_2 = 2^2 r_1 = 4(5.3\times 10^{-11}) = 2.12\times 10^{-10}\,\text{m}$$

For $n=3$:
$$r_3 = 3^2 r_1 = 9(5.3\times 10^{-11}) = 4.77\times 10^{-10}\,\text{m}$$

The maximum excitation energy available is $12.5\,\text{eV}$. In hydrogen, the energy needed to excite from $n=1$ to $n=2$ is $10.2\,\text{eV}$, but to $n=3$ is $12.09\,\text{eV}$, and to $n=4$ is $12.75\,\text{eV}$.

So the atom can be excited up to $n=3$, but not to $n=4$.

When it returns to lower levels, the possible emitted lines include transitions ending at $n=2$ and $n=1$. The visible lines correspond to transitions ending at $n=2$, which are the Balmer series.

Using Bohr-style quantisation for a circular orbit, the angular momentum is

$$L = n\frac{h}{2\pi}$$

For Earth's orbit,

$$L = mvr$$

Given:

• $m = 6.0\times10^{24}\,\text{kg}$
• $v = 3.0\times10^{4}\,\text{m/s}$
• $r = 1.5\times10^{11}\,\text{m}$
• $h = 6.6\times10^{-34}\,\text{J s}$

So,

$$n = \frac{2\pi mvr}{h}$$

First calculate $mvr$:

$$mvr = (6.0\times10^{24})(3.0\times10^{4})(1.5\times10^{11})$$
$$= 27.0\times10^{39} = 2.7\times10^{40}$$

Now,

$$n = \frac{2\pi(2.7\times10^{40})}{6.6\times10^{-34}}$$
$$\approx \frac{1.70\times10^{41}}{6.6\times10^{-34}}$$
$$\approx 2.6\times10^{74}$$

So the quantum number is of the order of $10^{74}$.

If we use the textbook's simplified constants more approximately, the result is still an extremely large number. The calculated value is $\approx 2.6\times10^{74}$.