Threshold frequency of a metal is $f_{0}$ . when light of frequency $v = 2 f_{0}$ is incident on the metal plate, the maximum velocity of the electron that is emitted is $v_{1}$ . When the frequency of incident radiation is $5 f_{0}$ , and the maximum velocity of the emitted electron is $v_{2}$ , find the ratio $\frac{v_{1}}{v_{2}}$ .

1 : 4

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1 : 2

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1 : 1

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1 : 32

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Solution

The correct option is B 1 : 2
K.E. of ejected electron is given by the equation,
$\frac{1}{2} m v^{2} = h (ν - ν_{o})$ , where $ν_{o}$ is threshold frequncy and $ν$ is incident frequency.
i) when $ν = 2 f_{o}$
$\frac{1}{2} m v_{1}^{2} = h (2 f_{o} - f_{o}) = h f_{o} . . .1$
ii) when $ν = 5 f_{o}$
$\frac{1}{2} m v_{2}^{2} = h (5 f_{o} - f_{o}) = h 4 f_{o} . .2$

From the $e q^{n} 1$ and $e q^{n} 2$ , we get
$\frac{v_{1}}{v_{2}} = 1 : 2$

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Similar questions

Q. Threshold frequency of a metal is

f_{0}

. when light of frequency

v = 2 f_{0}

is incident on the metal plate, the maximum velocity of the electron that is emitted is

v_{1}

. When the frequency of incident radiation is

5 f_{0}

, and the maximum velocity of the emitted electron is

v_{2}

, find the ratio

\frac{v_{1}}{v_{2}}

Threshold frequency of metal is $f_{0}$ . When light of frequency $θ$ = $2 f_{0}$ is incident on the metal plate, velocity of electron emitted in $V_{1}$ . When a frequency of incident radiation is $5 f_{0} . V_{2}$ is velocity of emitted electron, then $V_{1} : V_{2}$ is