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Dual Nature of Radiation and Matter

CBSE · Class 12 · Physics

NCERT Solutions for Dual Nature of Radiation and Matter — CBSE Class 12 Physics.

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11 Questions Solved · 1 Section

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EXERCISES

11.1Find the

(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.Show solution
For X-rays produced by electrons accelerated through V=30 kV=3.0×104 VV=30\,\text{kV}=3.0\times10^4\,\text{V}, the maximum photon energy is

Emax⁡=eV=(1.6×10−19)(3.0×104)=4.8×10−15 J.E_{\max}=eV=(1.6\times10^{-19})(3.0\times10^4)=4.8\times10^{-15}\,\text{J}.

(a) Maximum frequency:

νmax⁡=Emax⁡h=4.8×10−156.63×10−34≈7.25×1018 Hz.\nu_{\max}=\frac{E_{\max}}{h}=\frac{4.8\times10^{-15}}{6.63\times10^{-34}}\approx7.25\times10^{18}\,\text{Hz}.

(b) Minimum wavelength:

λmin⁡=cνmax⁡=3.0×1087.25×1018≈4.14×10−11 m.\lambda_{\min}=\frac{c}{\nu_{\max}}=\frac{3.0\times10^8}{7.25\times10^{18}}\approx4.14\times10^{-11}\,\text{m}.

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11.2The work function of caesium metal is 2.14 eV. When light of frequency 6 ×10¹⁴Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons,
(b) Stopping potential, and
(c) maximum speed of the emitted photoelectrons?Show solution
Use Einstein's photoelectric equation:

Kmax⁡=hν−ϕ0.K_{\max}=h\nu-\phi_0.

Given ν=6×1014 Hz\nu=6\times10^{14}\,\text{Hz},

hν=(6.63×10−34)(6×1014)=3.978×10−19 J.h\nu=(6.63\times10^{-34})(6\times10^{14})=3.978\times10^{-19}\,\text{J}.

In eV,

hν=3.978×10−191.602×10−19≈2.48 eV.h\nu=\frac{3.978\times10^{-19}}{1.602\times10^{-19}}\approx2.48\,\text{eV}.

Work function ϕ0=2.14 eV\phi_0=2.14\,\text{eV}.

So,

Kmax⁡=2.48−2.14=0.34 eV.K_{\max}=2.48-2.14=0.34\,\text{eV}.

Convert to joules:

Kmax⁡=0.34×1.602×10−19≈5.45×10−20 J.K_{\max}=0.34\times1.602\times10^{-19}\approx5.45\times10^{-20}\,\text{J}.

(b) Stopping potential:

eV0=Kmax⁡⇒V0=0.34 V.eV_0=K_{\max} \Rightarrow V_0=0.34\,\text{V}.

(c) Maximum speed:

Kmax⁡=12mv2,K_{\max}=\frac12 mv^2,

with me=9.11×10−31 kgm_e=9.11\times10^{-31}\,\text{kg},

v=2Kmax⁡m=2(5.45×10−20)9.11×10−31≈3.46×105 m/s.v=\sqrt{\frac{2K_{\max}}{m}}=\sqrt{\frac{2(5.45\times10^{-20})}{9.11\times10^{-31}}}\approx3.46\times10^5\,\text{m/s}.

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11.3The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?Show solution
The stopping potential is V0=1.5 VV_0=1.5\,\text{V}. From

Kmax⁡=eV0,K_{\max}=eV_0,

the maximum kinetic energy is

Kmax⁡=1.5 eV.K_{\max}=1.5\,\text{eV}.

In joules,

Kmax⁡=1.5×1.602×10−19=2.40×10−19 J.K_{\max}=1.5\times1.602\times10^{-19}=2.40\times10^{-19}\,\text{J}.

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11.4Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?Show solution
Given λ=632.8 nm=632.8×10−9 m\lambda=632.8\,\text{nm}=632.8\times10^{-9}\,\text{m} and P=9.42 mW=9.42×10−3 WP=9.42\,\text{mW}=9.42\times10^{-3}\,\text{W}.

### (a) Energy and momentum of each photon

Photon energy:

E=hcλE=\frac{hc}{\lambda}

E=(6.63×10−34)(3.0×108)632.8×10−9≈3.14×10−19 J.E=\frac{(6.63\times10^{-34})(3.0\times10^8)}{632.8\times10^{-9}}\approx3.14\times10^{-19}\,\text{J}.

Photon momentum:

p=Ec=hλp=\frac{E}{c}=\frac{h}{\lambda}

p=6.63×10−34632.8×10−9≈1.05×10−27 kg m/s.p=\frac{6.63\times10^{-34}}{632.8\times10^{-9}}\approx1.05\times10^{-27}\,\text{kg m/s}.

### (b) Photons per second

N=PE=9.42×10−33.14×10−19≈3.0×1016 photons/s.N=\frac{P}{E}=\frac{9.42\times10^{-3}}{3.14\times10^{-19}}\approx3.0\times10^{16}\,\text{photons/s}.

### (c) Speed of hydrogen atom with same momentum

For a hydrogen atom, p=mvp=mv.

Using mH≈1.67×10−27 kgm_H\approx1.67\times10^{-27}\,\text{kg},

v=pmH=1.05×10−271.67×10−27≈6.27×10−4 m/s.v=\frac{p}{m_H}=\frac{1.05\times10^{-27}}{1.67\times10^{-27}}\approx6.27\times10^{-4}\,\text{m/s}.

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11.5In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10⁻¹⁵ V s. Calculate the value of Planck's constant.Show solution
From Einstein's photoelectric equation,

V0=heν−ϕ0e,V_0=\frac{h}{e}\nu-\frac{\phi_0}{e},

so the slope of the V0V_0 vs ν\nu graph is

he.\frac{h}{e}.

Given slope =4.12×10−15 V s=4.12\times10^{-15}\,\text{V s},

h=e×slope=(1.602×10−19)(4.12×10−15)≈6.60×10−34 J s.h=e\times\text{slope}=(1.602\times10^{-19})(4.12\times10^{-15})\approx6.60\times10^{-34}\,\text{J s}.

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11.6The threshold frequency for a certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.Show solution
Using

eV0=h(ν−ν0),eV_0=h(\nu-\nu_0),

or in eV units,

V0=he(ν−ν0).V_0=\frac{h}{e}(\nu-\nu_0).

Here,

ν−ν0=(8.2−3.3)×1014=4.9×1014 Hz.\nu-\nu_0=(8.2-3.3)\times10^{14}=4.9\times10^{14}\,\text{Hz}.

Now

he≈4.14×10−15 V s,\frac{h}{e}\approx4.14\times10^{-15}\,\text{V s},

so

V0=(4.14×10−15)(4.9×1014)≈2.03 V.V_0=(4.14\times10^{-15})(4.9\times10^{14})\approx2.03\,\text{V}.

But the textbook uses the standard relation with precise values; using h=6.63×10−34 J sh=6.63\times10^{-34}\,\text{J s} and e=1.6×10−19 Ce=1.6\times10^{-19}\,\text{C} gives

V0=6.63×10−341.6×10−19×4.9×1014≈2.03 V.V_0=\frac{6.63\times10^{-34}}{1.6\times10^{-19}}\times4.9\times10^{14}\approx2.03\,\text{V}.

So the predicted cut-off voltage is **2.03 V2.03\,\text{V}**.

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11.7The work function for a certain metal is 4.2 eV. Will this metal give hotoelectric emission for incident radiation of wavelength 330 nm?
11.8Light of frequency 7.21 × 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵ m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
11.9Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
11.10What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 × 10⁻⁹ kg drifting with a speed of 2.2 m/s?
11.11Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

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