Question

The KE of photoelectron ejected by a metal upon irradiation with electromagnetic radiation of wavelength equal to that of the last line in Lyman series of He${}^{+}$ ion is: [I.P. of metal = 3.8 eV]

• A. 50.6 eV
• B. 52.3 eV
• C. 56.7 eV
• D. 59.0 eV

Answer 1

The correct option is A 50.6 eV

$K.E.=hv-h{v}_0=\frac{hc}{\lambda }-IP=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{\lambda }-3.8\times 1.6\times {10}^{-19}$

$\frac{1}{\lambda }=109700\times 2^2(\frac{1}{1^2}-0)$
So, $K.E.=50.6eV$.