Light of wavelength $500 n m$ is incident on a metal with a work function $2.28 e V$ . The de Borglie wavelength of the emitted electron is:

Question

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Answer 1

From photoelectric effect, $h ν = W_{0} + e V$

or $\frac{h c}{λ} = W_{0} + e V$

or $\frac{( 6 . 6 \times 1 0 ^{- 34} ) \times ( 3 \times 1 0 ^{8} )}{5 0 0 \times 1 0 ^{- 9}} = 2.28 \times 1.6 \times 1 0^{- 19} + e V$

or $e V = 0.31 \times 1 0^{- 19}$ or $V = 0.2$

Here V should be $V \leq 0.2$

The de Broglie wavelength of electron , $λ = \frac{1 2 . 2 7 \times 1 0 ^{- 10}}{V}$

so, $λ \geq 2.8 \times 1 0^{- 9} m$

Answer 2

From photoelectric effect,

h ν = W_{0} + e V

or

\frac{h c}{λ} = W_{0} + e V

or

\frac{( 6 . 6 \times 1 0 ^{- 34} ) \times ( 3 \times 1 0 ^{8} )}{5 0 0 \times 1 0 ^{- 9}} = 2.28 \times 1.6 \times 1 0^{- 19} + e V

or

e V = 0.31 \times 1 0^{- 19}

or

V = 0.2

Here V should be

V \leq 0.2

The de Broglie wavelength of electron ,

λ = \frac{1 2 . 2 7 \times 1 0 ^{- 10}}{V}

so,

λ \geq 2.8 \times 1 0^{- 9} m