42: Photoelectric Effect And Wave Particle Duality : Neet-xii-physics

Home Learn Fast select Course

Change Unit Change Subject Last Menu

NEET-XII-Physics

42: Photoelectric Effect and Wave Particle Duality

with Solutions - page 2

Note: Please signup/signin free to get personalized experience.

10 minutes can boost your percentage by 10%

Note: Please signup/signin free to get personalized experience.

Qstn #12
Consider the de-Broglie wavelength of an electron and a proton. Which wavelength is smaller if the two particles have
Ans : de-Broglie wavelength,
`` \lambda =\frac{h}{mv}``,
where h = Planck's constant
`` m`` = mass of the particle
v = velocity of the particle

#12-a
the same speed
Ans : It is given that the speed of an electron and proton are equal.
It is clear from the above equation that
`` \lambda \propto \frac{1}{m}``
As mass of a proton, m_p > mass of an electron, m_e, the proton will have a smaller wavelength compared to the electron.

#12-b
the same momentum
Ans : `` \lambda =\frac{h}{p}`` `` \left(\because p=mv\right)``
So, when the proton and the electron have same momentum, they will have the same wavelength.

#12-c
the same energy?
Ans : de-Broglie wavelength `` \left(\lambda \right)`` is also given by
`` \lambda =\frac{h}{\sqrt{2mE}}``,
where E = energy of the particle.
Let the energy of the proton and electron be E.
Wavelength of the proton,
`` {\lambda }_{p}=\frac{h}{\sqrt{2{m}_{\,\mathrm{\,p\,}}E}}...\left(1\right)``
Wavelength of the electron,
`` {\lambda }_{e}=\frac{h}{\sqrt{2{m}_{\,\mathrm{\,e\,}}E}}...\left(2\right)``
Dividing (2) by (1), we get:
`` \frac{{\lambda }_{e}}{{\lambda }_{p}}=\frac{\sqrt{{m}_{e}}}{\sqrt{{m}_{p}}}``
`` \Rightarrow \frac{{\lambda }_{e}}{{\lambda }_{p}}<1``
`` \Rightarrow {\lambda }_{e}<{\lambda }_{p}``
It is clear that the proton will have smaller wavelength compared to the electron.
Page No 363:

Qstn #13
If an electron has a wavelength, does it also have a colour?
Ans : Colour is a characteristic of electromagnetic waves. Electrons behave as a de-Broglie wave because of their velocity. A de-Broglie wave is not an electromagnetic wave and is one dimensional. Hence, no colour is shown by an electron.
Page No 363:

#
Section : ii

Qstn #1
Planck’s constant has the same dimensions as
(a) force × time
(b) force × distance
(c) force × speed
(d) force × distance time
digAnsr: d
Ans : (d) force × distance time
Planck's constant,
h = `` \frac{E}{v}=\frac{\,\mathrm{\,Force\,}\times \,\mathrm{\,distance\,}}{\,\mathrm{\,frequency\,}}``
`` \Rightarrow `` h = force `` \times `` distance time
Page No 363:

Qstn #2
Two photons of
(a) equal wavelength have equal linear momenta
(b) equal energies have equal linear momenta
(c) equal frequencies have equal linear momenta
(d) equal linear momenta have equal wavelengths
Ans : Two photons having equal linear momenta have equal wavelengths is correct. As in the rest of the options magnitude of momentum or energy can be same because energy and momentum are inversely proportional to wavelength. But the direction of propagation of the photons can be different.
Hence the correct option is D.
Page No 363:

Qstn #3
Let p and E denote the linear momentum and energy of a photon. If the wavelength is decreased,
(a) both p and E increase
(b) p increases and E decreases
(c) p decreases and E increases
(d) both p and E decrease
digAnsr: a
Ans : (a) both p and E increase
From the de-Broglie relation, wavelength,
`` \lambda =\frac{h}{p}...\left(1\right)``
`` \Rightarrow `` `` p=\frac{h}{\lambda }``
Here, h = Planck's constant
p = momentum of electron
It is clear from the above equation that `` p\propto \frac{1}{\lambda }``.
Thus, if the wavelength `` \left(\lambda \right)`` is decreased, then momentum `` \left(p\right)`` will be increase.
Relation between momentum and energy:
`` p=\sqrt{2mE}``
Here, E = energy of electron
m = mass of electron
Substituting the value of p in equation (1), we get:
`` \lambda =\frac{h}{\sqrt{2mE}}``
`` \Rightarrow \sqrt{E}=\frac{h}{\lambda \sqrt{2m}}``
`` \Rightarrow E=\frac{{h}^{2}}{2m{\lambda }^{2}}``
`` ``
Thus, on decreasing `` \lambda ``, the energy will increase.
Page No 363:

Qstn #4
Let n_r and n_b be the number of photons emitted by a red bulb and a blue bulb, respectively, of equal power in a given time.
(a) n_r = n_b
(b) n_r < n_b
(c) n_r > n_b
(d) The information is insufficient to derive a relation between n_r and n_b.
digAnsr: c
Ans : (c) n_r > n_b
The two bulbs are of equal power. It means that they consume equal amount of energy per unit time.
Now, as the frequency of blue light `` \left({f}_{b}\right)`` is higher than the frequency of red light`` \left({f}_{r}\right)``, `` h{f}_{b}>h{f}_{r}``.
Hence, the energy of a photon of blue light is more than the energy of a photon of red light.
Thus, a photon of blue light requires more energy than a photon of red light to be emitted.
For the same energy given to the bulbs in a certain time, the number of photons of blue light will be less than that of red light.
`` \therefore `` n_r > n_b (As the amount of energy emitted from the two bulb is same)
Page No 363:

Qstn #5
The equation E = pc is valid
(a) for an electron as well as for a photon
(b) for an electron but not for a photon
(c) for a photon but not for an electron
(d) neither for an electron nor for a photon
digAnsr: c
Ans : (c) for a photon but not for an electron
The equation E = pc is valid for a particle with zero rest mass. The rest mass of a photon is zero, but the rest mass of an electron is not zero. So, the equation will be valid for photon, and not electron.
Page No 363:

Qstn #6
The work function of a metal is hv₀. Light of frequency v falls on this metal. Photoelectric effect will take place only if
(a) v ≥ v₀
(b) v > 2v₀
(c) v < v₀
(d) v < v₀/2
digAnsr: a
Ans : (a) v ≥ v₀
As the work function of the metal is hv₀, the threshold frequency of the metal is v₀.
For photoelectric effect to occur, the frequency of the incident light should be greater than or equal to the threshlod frequency of the metal on which light is incident.
Page No 363:

Qstn #7
Light of wavelength λ falls on a metal with work-function hc/λ₀. Photoelectric effect will take place only if
(a) λ ≥ λ₀
(b) λ ≥ 2λ₀
(c) λ ≤ λ₀
(d) λ < λ₀/2
digAnsr: c
Ans : (c) λ ≤ λ₀
As the work-function of the metal is hc/λ₀, its threshold wavelength is λ₀.
For photoelectric effect, the wavelength of the incident light should be less than or equal to the threshold wavelength of the metal on which light is incident.
Page No 364:

Qstn #8
When stopping potential is applied in an experiment on photoelectric effect, no photoelectric is observed. This means that
(a) the emission of photoelectrons is stopped
(b) the photoelectrons are emitted but are re-absorbed
(c) the photoelectrons are accumulated near the collector plate
(d) the photoelectrons are dispersed from the sides of the apparatus
digAnsr: b
Ans : (b) the photoelectrons are emitted but are re-absorbed by the emitter metal
In an experiment on photoelectric effect, the photons incident at the metal plate cause photoelectrons to be emitted. The metal plate is termed as "emitter". The electrons ejected are collected at the other metal plate called "collector". When the potential of the collector is made negative with respect to the emitter (or the stopping potential is applied), the electrons emitted from the emitter are repelled by the collector. As a result, some electrons go back to the cathode and the current decreases.
Page No 364:

Qstn #9
If the frequency of light in a photoelectric experiment is doubled, the stopping potential will
(a) be doubled
(b) be halved
(c) become more than double
(d) become less than double
digAnsr: c
Ans : (c) become more than double
According to Einstein's equation of photoelectric effect,
`` e{V}_{0}=hv-\phi ``
`` \Rightarrow {V}_{0}=\frac{hv-\phi }{e}...\left(1\right)``
Here, V₀ = stopping potential
v = frequency of light
`` \phi `` = work function
Let the new frequency of light be 2ν and the corresponding stopping potential be V₀'.
Therefore,
`` e{V}_{0}\text{'}=2hv-\phi ``
`` {V}_{0}\text{'}=\frac{2hv-\phi }{e}...\left(2\right)``
Multiplying both sides of equation (1) by 2, we get:
`` 2{V}_{0}=\frac{2hv-2\phi }{e}...\left(3\right)``
Now if we compare (2) and (3), it can be observed that:
`` \frac{2hv-\phi }{e}>\frac{2hv-2\phi }{e}``
`` \Rightarrow {V}_{0}\text{'}>2{V}_{0}``
It is clear from the above equation that if the frequency of light in a photoelectric experiment is doubled, the stopping potential will be more than doubled.
Page No 364: