## From the above figure the values of stopping potentials for M1 and M2 for a frequency v3 (> vo2) ….

Q: From the above figure the values of stopping potentials for M1 and M2 for a frequency v3 (> v02) of the incident radiations are V1 and V2 respectively. Then the slope of the line is equal to

(a) $\displaystyle \frac{V_2 -V_1}{\nu_{02} - \nu_{01}}$

(b) $\displaystyle \frac{V_1 -V_2}{\nu_{02} - \nu_{01}}$

(c) $\displaystyle \frac{V_2}{\nu_{02} - \nu_{01}}$

(d) $\displaystyle \frac{V_1}{\nu_{02} - \nu_{01}}$

Ans: (a)

Sol:
$\displaystyle h\nu_{01} = \phi + eV_1$

$\displaystyle V_1 = \frac{h \nu_{01}}{e} -\frac{\phi}{e}$ …(i)

Similarly ,

$\displaystyle V_2 = \frac{h \nu_{02}}{e} -\frac{\phi}{e}$ …(ii)

subtracting (i) from(ii)

$\displaystyle V_2 - V_1 = \frac{h}{e}(\nu_{02}-\nu_{01})$

$\displaystyle \frac{h}{e} = \frac{V_2 -V_1}{\nu_{02} - \nu_{01}}$

## From the graph shown, the value of Work function if the stopping potential (V)…

Q: From the graph shown, the value of Work function if the stopping potential (V), and frequency of the incident light, v, are on y and x-axes respectively is

(a) 1eV

(b) 2eV

(c) 4eV

(d) 3eV

Ans:(b)

Sol:
$\displaystyle h\nu = \phi + K_{max}$

$\displaystyle h\nu = h\nu_0 + eV$

$\displaystyle V = \frac{h\nu}{e} - \frac{h\nu_0}{e}$

On comparing with y = mx + c

Slope = h/e , Intercept = hν0/e

$\displaystyle \frac{h\nu_0}{e} = 2$

0 = 2eV

φ0 = 2eV

## An electron is accelerated through a potential difference of V volt. It has a wavelength λ associated with it…..

Q. An electron is accelerated through a potential difference of V volt. It has a wavelength λ associated with it. Through what potential difference an electron must be accelerated so that its de Broglie wavelength is the same as that of a proton? Take mass of proton to be 1837 times larger than the mass of electron.

(a) V volt

(b) 1837V volt

(c) V/1837 volt

(d) √(1837)volt

Ans: (d)

Sol: λ = h/√(2meV)

λ’ = h/√(2m’eV’)

Since , λ = λ’

h/√(2meV) = h/√(2m’eV’)

m V =m’V’

m V = m×1837×V’

V’ = V/1837 volt

## When a metallic surface is illuminated by a light of frequency 8×10^14 Hz, photoelectron of maximum energy …..

Q. When a metallic surface is illuminated by a light of frequency 8×1014 Hz, photoelectron of maximum energy 0.5eV is emitted. When the same surface is illuminated by light of frequency 12 × 1014 Hz, photoelectron of maximum energy 2eV is emitted. The work function is

(a) 0.5 eV

(b) 2.85 eV

(c) 2.5 eV

(d) 3.5 eV

Ans: (c)

Sol: In first case ,

h(8×1014) = φ+0.5    ….(i)

In 2nd case ,

h(12 × 1014) = φ+2  ….(ii)

Dividing (ii) by (i)

h(12 × 1014) /h(8×1014)= (φ+2)/(φ+0.5)

3/2 =  (φ+2)/(φ+0.5)

2φ + 4 = 3φ + 1.5

φ = 2.5 eV

## A modern 200W sodium street lamp emits yellow light of wavelength 0.6mm. Assuming it to be 25% efficient …..

Q. A modern 200W sodium street lamp emits yellow light of wavelength 0.6mm. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is

(a) 62 × 1020

(b) 3 × 1019

(c) 1.5 × 1020

(d) 6 × 1018

Ans: (c)

Sol: Energy of photon , E= hc/λ

E = 6.6×10-34 × 3× 108/(0.6× 10-6)

E = 3.3× 10-19 J

Efficiency , η = P0/PI

25/100 = P0/200

P0 = 50 W

Number of photons emitted per sec

N = P0/E = 50/3.3× 10-19 = 1.5× 1020