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When a beam of $10.6 eV$ photons of intensity 2Wm^-2 falls on a platinum surface of area $1.0 \times 10^{- 4} m^{2}$ and work function $5.6 eV,$ $53 %$ of the incident photon eject photoelectrons. Find the number of photoelectrons emitted per second and their minimum and maximum energies (in eV).

When a beam of 10.6 eV photons of intensity……. :-

Solution.

Intensity is given by

$\displaystyle I=\frac{Energy(E)}{Area(A)\times time(t)}=\frac{N(h\nu )}{A\times t}$

Where N is number of photons incident in t time.

$\displaystyle \Rightarrow \frac{N}{t}=\frac{IA}{hv}$

As $53 %$ of the incident photon eject photoelectrons, so

Number of photoelectrons emitted per sec $\displaystyle \frac{N}{t}=\frac{IA}{hv}\times \frac{0.53}{100}$

$\displaystyle \frac{N}{t}=\frac{2\times (1.0\times {{10}^{-4}})}{10.6\times 1.6\times {{10}^{-19}}}\times \frac{0.53}{100}=6.25\times {{10}^{11}}$

Minimum kinetic energy of photoelectrons, K_min = 0 and

Maximum kinetic energy of photoelectrons,

K_max = E – Φ₀ = (10.6 – 5.6)eV = 5 eV