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Standard XII

Physics

Stopping Potential

For photoelec...

Question

For photoelectric emission from certain metal the cut-off frequency is $v$ . If radiation of frequency $2 v$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be:
( $m$ is the electron mass)

\sqrt{\frac{h v}{(2 m)}}

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\sqrt{\frac{h v}{m}}

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\sqrt{\frac{2 h v}{m}}

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2 \sqrt{\frac{h v}{m}}

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Solution

The correct option is D $\sqrt{\frac{2 h v}{m}}$
K.E $= E - ϕ$

$\frac{1}{2} m v_{m a x}^{2} = h 2 v - h v$

$v_{m a x} = \sqrt{\frac{2 h v}{m}}$

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For photoelectric emission from certain metal the cut-off frequency is v. If radiation of frequency 2v impinges on the metal plate, the maximum possible velocity of the emitted electron will be: (m is the electron mass)

The correct option is D √2hvmK.E =E−ϕ12mv2max=h2v−hvvmax=√2hvm

For photoelectric emission from certain metal the cut-off frequency is $v$ . If radiation of frequency $2 v$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be:
( $m$ is the electron mass)

The correct option is D $\sqrt{\frac{2 h v}{m}}$
K.E $= E - ϕ$

$\frac{1}{2} m v_{m a x}^{2} = h 2 v - h v$

$v_{m a x} = \sqrt{\frac{2 h v}{m}}$