Maximum velocity of the photoelectron emitted by a metal is 1.8 × 106 ms-1. Take the value of specific charge of the electron is 1.8 ×1011 C kg-1. Then the stopping potential in volt is

Question

Maximum velocity of the photoelectron emitted by a metal is 1.8 × 106 ms-1. Take the value of specific charge of the electron is 1.8 ×1011 C kg-1. Then the stopping potential in volt is (A) 1 (B) 3 (C) 9 (D) 6

Tardigrade Admin · Accepted Answer

Given, v=1.8 × 106 m / s ; (e/m)=1.8 × 1011 C / ,kg We have, e V0=(1/2) m v2 ⇒ V0 (e/m)=(v2/2) ⇒ V0 × 1.8 × 1011=(1.8 × 1.8 ×(106)2/2) ⇒ V0=9 V

Q. Maximum velocity of the photoelectron emitted by a metal is $1.8 \times 1 0^{6} m s^{- 1}$ . Take the value of specific charge of the electron is $1.8 \times 1 0^{11} C k g^{- 1}$ . Then the stopping potential in volt is

1

3

9

6

Given, $v = 1.8 \times 1 0^{6} m / s; \frac{e}{m} = 1.8 \times 1 0^{11} C /, k g$
We have, $e V_{0} = \frac{1}{2} m v^{2}$
$\Rightarrow V_{0} \frac{e}{m} = \frac{v ^{2}}{2}$
$\Rightarrow V_{0} \times 1.8 \times 1 0^{11} = \frac{1.8 \times 1.8 \times ( 1 0 ^{6} ) ^{2}}{2}$
$\Rightarrow V_{0} = 9 V$

Q. Maximum velocity of the photoelectron emitted by a metal is 1.8×106ms−1. Take the value of specific charge of the electron is 1.8×1011Ckg−1. Then the stopping potential in volt is

1

3

9

6

Given, v=1.8×106m/s;me​=1.8×1011C/,kg We have, eV0​=21​mv2 ⇒V0​me​=2v2​ ⇒V0​×1.8×1011=21.8×1.8×(106)2​ ⇒V0​=9V

Q. Maximum velocity of the photoelectron emitted by a metal is $1.8 \times 1 0^{6} m s^{- 1}$ . Take the value of specific charge of the electron is $1.8 \times 1 0^{11} C k g^{- 1}$ . Then the stopping potential in volt is

Given, $v = 1.8 \times 1 0^{6} m / s; \frac{e}{m} = 1.8 \times 1 0^{11} C /, k g$
We have, $e V_{0} = \frac{1}{2} m v^{2}$
$\Rightarrow V_{0} \frac{e}{m} = \frac{v ^{2}}{2}$
$\Rightarrow V_{0} \times 1.8 \times 1 0^{11} = \frac{1.8 \times 1.8 \times ( 1 0 ^{6} ) ^{2}}{2}$
$\Rightarrow V_{0} = 9 V$