Photons of energy $1 e V$ and $2.5 e V$ successively illuminated a metal whose work function is $0.5 e V$ . The ratio of maximum speeds of emitted electron is______________.

1 : 3

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1 : 2

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3 : 1

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2 : 1

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Solution

The correct option is B $1 : 2$
Energy of photon $E = ϕ + \frac{1}{2} m v^{2}$
$v_{m a x} = \sqrt{\frac{2 (E - ϕ)}{m}}$
So putting vales
$\frac{v_{m a x 1}}{v_{m a x 2}} = \sqrt{\frac{1 - 0.5}{2.5 - 0.5}}$

$\frac{v_{m a x 1}}{v_{m a x 2}} = \frac{1}{2}$

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Photons of energy 1eV and 2.5eV successively illuminated a metal whose work function is 0.5eV. The ratio of maximum speeds of emitted electron is______________.

The correct option is B 1:2Energy of photon E=ϕ+12mv2vmax=√2(E−ϕ)mSo putting valesvmax1vmax2=√1−0.52.5−0.5vmax1vmax2=12

Photons of energy $1 e V$ and $2.5 e V$ successively illuminated a metal whose work function is $0.5 e V$ . The ratio of maximum speeds of emitted electron is______________.

The correct option is B $1 : 2$
Energy of photon $E = ϕ + \frac{1}{2} m v^{2}$
$v_{m a x} = \sqrt{\frac{2 (E - ϕ)}{m}}$
So putting vales
$\frac{v_{m a x 1}}{v_{m a x 2}} = \sqrt{\frac{1 - 0.5}{2.5 - 0.5}}$

$\frac{v_{m a x 1}}{v_{m a x 2}} = \frac{1}{2}$