Photon having the energy equivalent to the binding energy of 4th state of $H e^{+}$ atom is used to eject an electron from the metal surface of work function 1.4 eV. If electrons are further accelerated through the potential difference of 4 V then determine the minimum value of de-Broglie wavelength associated with the electron.

5 A^{o}

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2.5 A^{o}

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7.6 A^{o}

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10 A^{o}

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Solution

The correct option is A $5 A^{o}$
Energy of Photon = $13.6 \times \frac{2^{2}}{4^{2}} = 3.4 e V$

Energy left after overcoming work function = $(3.4 - 1.4) = 2 e V$

Energy of accelerated photon = $(2 + 4) = 6 e V$

De-broglie wavelength $λ = \frac{h}{\sqrt{2 m E}}$

$= \frac{6.62 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 6 \times 1.6 \times 10^{- 19}}}$

$= 5 \times 10^{- 10} m$ $= 5 A^{o}$

Hence, the correct option is $A$

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