Photoelectric Effect

Q: When a centimeter thick surface is illuminated with light of wavelength λ , the stopping potential is V. When the same surface is illuminated by light of wavelength 2λ , the stopping potential is V/3. Threshold wavelength for the metallic surface is

(A) 4λ/3

(B) 4λ

(C) 6λ

(D) 8λ/3

Solution :

$\large \frac{hc}{\lambda} = \phi + e V $

$\large \frac{hc}{\lambda} – \phi = e V $ …(i)

$\large \frac{hc}{2 \lambda} = \phi + e \frac{V}{3} $

$\large \frac{hc}{2 \lambda} – \phi = e \frac{V}{3} $ …(ii)

On dividing (i) by (ii)

$\large \frac{\frac{hc}{\lambda} – \phi }{\frac{hc}{2 \lambda} – \phi } = 3 $

$\large \frac{hc}{\lambda} – \phi = 3 (\frac{hc}{2 \lambda} – \phi) $

$\large 3 \phi – \phi = \frac{3 hc}{2 \lambda} – \frac{hc}{\lambda} $

$\large 2 \phi = \frac{hc}{2 \lambda} $

$\large \phi = \frac{hc}{4 \lambda} $

$\large \frac{hc}{ \lambda_o} = \frac{hc}{4 \lambda} $

$\large \lambda_o = 4 \lambda$

Correct option is (B)

Q: If h is Planck’s constant is SI system, the momentum of a photon of wavelength 0.01 A° is:

(A) 10^–2 h

(B) h

(C) 10²

(D) 10¹²h

Solution : $\large \lambda = \frac{h}{p}$

$\large p = \frac{h}{\lambda}$

$\large p = \frac{h}{0.01 \times 10^{-10}}$

p = 10¹²h

Q: An electron with initial kinetic energy of 100 eV is acceleration through a potential difference of 50 V. Now the de-Broglie wavelength of electron becomes.

(A) 1 A°

(B) $\sqrt{1.5}$ A°

(C) $\sqrt{3}$ A°

(D) 12.27 A°

Solution : Kinetic Energy , K = 100 eV + 50 eV = 150 e V

$\large \lambda = \sqrt{\frac{150}{V}}$

$\large \lambda = \sqrt{\frac{150}{150}}$

= 1 A°

Correct option is (A)

Q: A proton and an electron are accelerated by same potential difference have de-Broglie wavelength λ_p and λ_e

(A) λ_e = λ_p

(B) λ_e < λ_p

(C) λ_e > λ_p

(D) none of these

Solution :

As Proton & electron have same charge and accelerated by same Potential difference V . Therefore Kinetic Energy K = q V will be constant .

De-broglie Wavelength $\large \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m K}}$

$\large \lambda \propto \frac{1}{\sqrt{m}}$

$\large \frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m_e}{m_p} }$

$\large \frac{\lambda_p}{\lambda_e} < 1 $ (Since m_p > m_e)

Hence , λ_e > λ_p

Correct option is (C)

Q: Photons with energy 5eV are incident on a cathode C, on a photoelectric cell. The maximum energy of the 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

(A) 3 V

(B) –3 V

(C) –1 V

(D) 4 V

Solution : KE_max = 2 ev

E_photon = 5ev

φ = 5 ev – 2 ev = 3 ev

Now no current when

E_photon = 6ev

i.e. KE_max  < 3ev

eV_max < 3ev

V_max < 3ev/e = 3V max

φ = 3eV

K.E. = 3eV in Second case

Correct option is (B)

Q: Cut off potentials for a metal in photoelectric effect for light of wavelength λ₁ , λ₂ and λ₃ is found to be V₁ , V₂ and V₃ volts if V₁ , V₂ and V₃ are in Arithmetic Progression then λ₁ , λ₂ and λ₃ will be :

(A) Arithmetic Progression

(B) Geometric Progression

(C) Harmonic Progression

(D) None

Solution :

$\large \frac{hc}{\lambda_1} = \phi + e V_1 $

$\large \frac{hc}{\lambda_2} = \phi + e V_2 $

$\large \frac{hc}{\lambda_3} = \phi + e V_3 $

As V₁ , V₂ and V₃ are in Arithmetic Progression .

φ + e V₁ , φ + e V₂ and φ + e V₃ are also in Arithmetic Progression .

$\large \frac{hc}{\lambda_1} , \frac{hc}{\lambda_2} , \frac{hc}{\lambda_3} $ are in A.P

$\large \frac{1}{\lambda_1} , \frac{1}{\lambda_2} , \frac{1}{\lambda_3} $ are in A.P

Therefore , λ₁ , λ₂ and λ₃ will be in H.P

correct option is (C)

Q: A point source of light is used in photoelectric effect. If the source is removed farther from the emitting metal, the stopping potential :

(A) will increase

(B) will decrease

(C) will remain constant

(D) will either increase or decrease

Solution : Depends on frequency not on Intensity .

Correct option is (C)

Q: 10^–3 W of 5000 A° light is directed on a photoelectric cell. If the current in the cell is 0.16 mA , the percentage of incident photons which produce Photoelectrons is

(A) 0.4 %

(B) 0.04 %

(C) 20 %

(D) 10

Solution : $\large E = \frac{h c}{\lambda } $

$\large E = \frac{12400}{5000} eV = \frac{12400}{5000} \times 1.6 \times 10^{-19} J$

No. of Photon $\large N = \frac{P}{E} = \frac{10^{-3}}{\frac{12400}{5000} \times 1.6 \times 10^{-19}}$

N = 0.25 x 10¹⁶

No. of Electron reaching $\large = \frac{0.16 \times 10^{-6}}{1.6 \times 10^{-19}} $

= 10⁺¹²

Percentage $\large = \frac{10^{12}}{0.25 \times 10^{16}} \times 100 $

= 0.04 %

Correct option is (B)

Q: Let n_r and n_b be respectively the number of photons emitted by a red bulb and a blue bulb of equal power in a given time.

(A) n_r = n_b

(B) n_r < n_b

(C) n_r > n_b

(D) data insufficient

Solution : $\large E = \frac{12400}{\lambda (A^o)}$

Number of Photon $\large N = \frac{P}{E} = \frac{P}{hc/\lambda}$

$\large N = \frac{P \lambda}{h c} $

$\large N \propto \lambda $

Correct option is (C)

Q: When a photon of light collides with a metal surface , number of electrons, (if any) coming out is

(A) only one

(B) only two

(C) infinite

(D) depends upon factors

Solution : A Photon can interact with only a single electron.

Correct option is (A)