Maximum velocity of photoelectron emitted is 4.8 ms -1. The (e/M) ratio of electron is 1.76 × 1011 Ckg -1, then stopping potential is given by

Question

Maximum velocity of photoelectron emitted is 4.8 ms -1. The (e/M) ratio of electron is 1.76 × 1011 Ckg -1, then stopping potential is given by (A) 5 × 10-10 JC -1 (B) 3 × 10-7 JC -1 (C) 7 × 1011 JC -1 (D) 2.5 × 10-2 JC -1

Tardigrade Admin · Accepted Answer

As, eV s=(1/2) m vm2 or Vs=(m vm2/2 e)=(vm2/2(e / m)) =((4.8)2/2 × 1.76 × 10-11)=7 × 1011 JC -1

Q. Maximum velocity of photoelectron emitted is $4.8 m s^{- 1}$ . The $\frac{e}{M}$ ratio of electron is $1.76 \times 1 0^{11} C k g^{- 1}$ , then stopping potential is given by

$5 \times 1 0^{- 10} J C^{- 1}$

$3 \times 1 0^{- 7} J C^{- 1}$

$7 \times 1 0^{11} J C^{- 1}$

$2.5 \times 1 0^{- 2} J C^{- 1}$

As, $e V_{s} = \frac{1}{2} m v_{m}^{2}$
or $V_{s} = \frac{m v _{m}^{2}}{2 e} = \frac{v _{m}^{2}}{2 ( e / m )}$
$= \frac{( 4.8 ) ^{2}}{2 \times 1.76 \times 1 0 ^{- 11}} = 7 \times 1 0^{11} J C^{- 1}$

Q. Maximum velocity of photoelectron emitted is 4.8ms−1. The Me​ ratio of electron is 1.76×1011Ckg−1, then stopping potential is given by

5×10−10JC−1

3×10−7JC−1

7×1011JC−1

2.5×10−2JC−1