Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is [2015]

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Solution

Q.
Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is [2015]

1 $\geq 2.8 \times 10^{- 9} m$

2 $\leq 2.8 \times 10^{- 12} m$

3 $< 2.8 \times 10^{- 10} m$

4 $< 2.8 \times 10^{- 9} m$

Ans.
(1)

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted electron is

   $K_{max} = \frac{h c}{λ} - ϕ_{0}$

where $λ$ is the wavelength of incident light and $ϕ_{0}$ is the work function.

Here, $λ = 500$ $nm, h c = 1240$ $eV$ $nm, ϕ_{0} = 2.28$ $eV$

$∴ K_{max} = \frac{1240 eV nm}{500 nm} - 2.28$ $eV$

   $= 2.48 eV - 2.28 eV = 0.2 eV$

The de Broglie wavelength of the emitted electron is

$λ_{min} = \frac{h}{\sqrt{2 m K_{max}}}$

where $h$ is the Planck's constant and $m$ is the mass of the electron.

As $h = 6.6 \times 10^{- 34} J s, m = 9 \times 10^{- 31} kg$

and $K_{max} = 0.2 eV = 0.2 \times 1.6 \times 10^{- 19} J$

$∴$    $λ_{min} = \frac{6.6 \times 10^{- 34} J s}{\sqrt{2 (9 \times 10^{- 31} kg) (0.2 \times 1.6 \times 10^{- 19} J)}}$

$= \frac{6.6}{2.4} \times 10^{- 9} m \approx 2.8 \times 10^{- 9} m$

so,   $λ \geq 2.8 \times 10^{- 9} m$

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