Question

Light of wavelength 4000 Aº falls on the surface
of cesium. Calculate the energy of the photo-
electron emitted. The critical wavelength for
photoelectric effect in cesium is 6600 A.​

Answer 1

Answer:

Explanation:

First of all, you must have knowledge of the following formulas.

wavelegnth = c/frequency

where c = speed of light in space

frequency = $v$

energy of photoelectron emitted = $hv - hv_{0}$

where $v_{0}$ = threshhold/critical frequency

let the symbol for wave legnth = $\lambda$

$energy = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}}$

h = $6.626 \times 10^{-34}$

c = $3 \times 10^{8}$

after solving we get  $1.95 \times 10^{-19} J$

or 1.218 eV Energy

Thanks!