Lithium has work function of $2.3 e V$ . It is exposed to light of wavelength $4.8 \times 10^{- 7} m$ . Find the maximum kinetic energy with which electron leaves the surface. Also calculate the longest wavelength which can produce the photo electrons.

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Solution

Wavelength $λ = 480 \times 10^{- 9} m = 480 n m$
So energy of incident photon= $\frac{h c}{λ} = \frac{1237.5}{480 n m} e v = 2.58 e v$ or equal to $2.6 e v$

Maximum kinetic energy $= I n c i d e n t - W . f u n c t i o n = 2.6 - 2.3 = 0.3 e v$
Longest wavelength means minimum energy which will be just equal to work function means the kinetic energy of emitted electron willbe zero.
$λ_{m a x} = \frac{237.5}{E} = \frac{237.5}{2.3} n m = 103.2 n m$

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Lithium has work function of 2.3 eV. It is exposed to light of wavelength 4.8×10−7 m. Find the maximum kinetic energy with which electron leaves the surface. Also calculate the longest wavelength which can produce the photo electrons.

Lithium has work function of $2.3 e V$ . It is exposed to light of wavelength $4.8 \times 10^{- 7} m$ . Find the maximum kinetic energy with which electron leaves the surface. Also calculate the longest wavelength which can produce the photo electrons.