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- 1011C kg-1- Then the stopping potential in volt is-

The correct option is B 9 given : #160; #160; v = 1.8× 10^6 #160; m/s , #160; e/m = 1.8× 10^11 #160; C/kg #160; ...

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Standard XII

Physics

Stopping Potential

Maximum veloc...

Question

Maximum velocity of the photoelectron emitted by a metal is $1.8 \times 10^{6} m s^{- 1}$ . Take the value of specific charge of the electron is $1.8 \times 10^{11} C k g^{- 1}$ . Then the stopping potential in volt is:

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Solution

The correct option is B $9$
given : $v = 1.8 \times 10^{6}$ $m / s$ , $e / m = 1.8 \times 10^{11}$ $C / k g$
Stopping potential $V = \frac{K . E}{e} = \frac{\frac{1}{2} m v^{2}}{e}$
$∴$ $V = \frac{(1.8 \times 10^{6})^{2}}{2 \times 1.8 \times 10^{11}} = 9$ volts

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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:

The correct option is B 9given : v=1.8×106 m/s , e/m=1.8×1011 C/kg Stopping potential V=K.Ee=12mv2e∴ V=(1.8×106)22×1.8×1011=9 volts

Maximum velocity of the photoelectron emitted by a metal is $1.8 \times 10^{6} m s^{- 1}$ . Take the value of specific charge of the electron is $1.8 \times 10^{11} C k g^{- 1}$ . Then the stopping potential in volt is:

The correct option is B $9$
given : $v = 1.8 \times 10^{6}$ $m / s$ , $e / m = 1.8 \times 10^{11}$ $C / k g$
Stopping potential $V = \frac{K . E}{e} = \frac{\frac{1}{2} m v^{2}}{e}$
$∴$ $V = \frac{(1.8 \times 10^{6})^{2}}{2 \times 1.8 \times 10^{11}} = 9$ volts