Radiations of frequency 10¹⁵ Hz are incident on two photosensitive surfaces A and B. Following observations are recorded:
Surface A: No photoemission takes place.
Surface B: Photoemission takes place but photoelectrons have zero energy.
Explain the above observations on the basis of Einstein's photoelectric equation.
How will the observation with surface B change when the wavelength of incident radiations is decreased?

Solution

Given,
Radiation of frequencies incident on the photoelectric surface = 10¹⁵ Hz

Now, according to Einstein's photoelectric equation, kinetic energy of a photoelectron is given by,

K_{\max} = hv - ϕ_{0} \Rightarrow K_{\max} = h (v - v_{0})

i.e.,

K_{\max} = hc [\frac{1}{λ} - \frac{1}{λ_{0}}]

(i) The threshold frequency of surface A is greater than 10¹⁵ Hz, and that is the reason no photoemission takes place.

(ii) For surface B, the threshold frequency is equal to 10¹⁵ Hz. So, photoemission takes place but photoelectrons have zero kinetic energy.
If, the kinetic energy of the elctrons emitted from surface B has to be increased then, the wavelength of the incident radiation has to be decreased.

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