Modern Physics - NEET Physics Chapterwise MCQs & PYQs

NEET Modern Physics MCQs & PYQs

Practice NEET Modern Physics Questions

Question 1:

easy

The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of $3.3 \times 10^{-3}$ watt will be: ($h = 6.6 \times 10^{-34}$ Js)

(2021)

Energy of one photon is $E = \frac{hc}{\lambda}$. The number of photons emitted per second is $n = \frac{P}{E} = \frac{P \lambda}{hc}$. Substituting values: $n = \frac{3.3 \times 10^{-3} \times 600 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} = 10^{16}$.

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Question 2:

easy

Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?

(2020)

Initial frequency is $1.5\nu_0$. When the frequency is halved, the new frequency becomes $0.75\nu_0$. Since the new frequency is less than the threshold frequency $\nu_0$, no photoelectric emission will take place, regardless of the intensity. Thus, the photoelectric current will be zero.

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Question 3:

easy

When the light of frequency $2\nu_0$ (where $\nu_0$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5\nu_0$, the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is:

(2018)

Using Einstein's photoelectric equation, $h\nu = h\nu_0 + \frac{1}{2}mv^2$. For the first case, $h(2\nu_0) = h\nu_0 + \frac{1}{2}mv_1^2 \implies \frac{1}{2}mv_1^2 = h\nu_0$. For the second case, $h(5\nu_0) = h\nu_0 + \frac{1}{2}mv_2^2 \implies \frac{1}{2}mv_2^2 = 4h\nu_0$. Dividing the two equations gives $\frac{v_1^2}{v_2^2} = \frac{1}{4}$, which means $\frac{v_1}{v_2} = \frac{1}{2}$.

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Question 4:

easy

The photoelectric threshold wavelength of silver is $3250 \times 10^{-10}$ m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{-10}$ m is:
(Given $h = 4.14 \times 10^{-15}$ eV and $c = 3 \times 10^8$ ms$^{-1}$)

(2017-Delhi)

 

Work function $W = \frac{hc}{\lambda_0} = \frac{1242 eV nm}{325 nm} \approx 3.82 eV$. Energy of incident photon $E = \frac{hc}{\lambda} = \frac{1242 eV nm}{253.6 nm} \approx 4.89 eV$. Maximum kinetic energy $K_{max} = E - W = 4.89 - 3.82 = 1.07 eV$. $K_{max} = 1.07 \times 1.6 \times 10^{-19} J \approx 1.7 \times 10^{-19} J$. Also $K_{max} = \frac{1}{2}mv^2$, so $v = \sqrt{\frac{2 \times 1.7 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 0.61 \times 10^6 ms^{-1}$.

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Question 5:

easy

When a metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is V. If the same surface is illuminated with radiation of wavelength $2\lambda$, the stopping potential is V/4. The threshold wavelength for the metallic surface is:

(2016-I)

From photoelectric equation: $\frac{hc}{\lambda} = \frac{hc}{\lambda_0} + eV$ and $\frac{hc}{2\lambda} = \frac{hc}{\lambda_0} + \frac{eV}{4}$. Multiplying the second equation by 4 gives $\frac{2hc}{\lambda} = \frac{4hc}{\lambda_0} + eV$. Subtracting the first equation from this result yields $\frac{hc}{\lambda} = \frac{3hc}{\lambda_0}$, which simplifies to $\lambda_0 = 3\lambda$.

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Question 6:

easy

Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is:

(2016-II)

Work function $W = E_1 - K_1 = 5 eV - 2 eV = 3 eV$. For the second case, maximum kinetic energy $K_2 = E_2 - W = 6 eV - 3 eV = 3 eV$. The stopping potential $V_0 = \frac{K_2}{e} = 3 V$. Since the anode must be at a negative potential relative to the cathode to repel the electrons, the potential is -3 V.

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Question 7:

easy

A certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for photo-electric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$, the stopping potential is $V_0$. The threshold wavelength for this surface for photoelectric effect is:

(2015)

Using the photoelectric equation: $\frac{hc}{\lambda} = W + 3eV_0$ and $\frac{hc}{2\lambda} = W + eV_0$. Multiplying the second equation by 3 gives $\frac{3hc}{2\lambda} = 3W + 3eV_0$. Subtracting the first equation from this gives $\frac{hc}{2\lambda} = 2W$. Thus, $W = \frac{hc}{4\lambda}$. Since $W = \frac{hc}{\lambda_0}$, we find $\lambda_0 = 4\lambda$.

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Question 8:

easy

A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\lambda/2$. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is:
($h$ = Plank’s constant, $c$ = speed of light)

(2015 Re)

Let K be the initial kinetic energy. $\frac{hc}{\lambda} = W + K$ and $\frac{hc}{\lambda/2} = \frac{2hc}{\lambda} = W + 3K$. Multiplying the first equation by 3 gives $\frac{3hc}{\lambda} = 3W + 3K$. Subtracting the second equation from this gives $\frac{hc}{\lambda} = 2W$. Therefore, the work function is $W = \frac{hc}{2\lambda}$.

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Question 9:

easy

10. When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is:

(2014)

Let initial energy be $E$. We have $E = W + 0.5$. When energy is increased by 20%, new energy is $1.2E$. So, $1.2E = W + 0.8$. Substituting $E = W + 0.5$ into the second equation gives $1.2(W + 0.5) = W + 0.8$. This expands to $1.2W + 0.6 = W + 0.8$, yielding $0.2W = 0.2$, which means $W = 1.0 eV$.

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Question 10:

easy

17. Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be: (2011 Pre)

Maximum kinetic energy $K_{max} = E - W$. For the first light, $K_1 = 1 - 0.5 = 0.5 eV$. For the second light, $K_2 = 2.5 - 0.5 = 2.0 eV$. The ratio of their kinetic energies is $K_1/K_2 = 1/4$. The ratio of maximum speeds is $v_1/v_2 = \sqrt{K_1/K_2} = \sqrt{1/4} = 1/2$.

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