JEE MAIN - Physics (2022 - 28th June Evening Shift - No. 17)
Let K1 and K2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $$\lambda$$1 and $$\lambda$$2, respectively are incident on a metallic surface. If $$\lambda$$1 = 3$$\lambda$$2 then :
$${K_1} > {{{K_2}} \over 3}$$
$${K_1} < {{{K_2}} \over 3}$$
$${K_1} = {{{K_2}} \over 3}$$
$${K_2} = {{{K_1}} \over 3}$$
Explanation
$${K_1} = {{hc} \over {{\lambda _1}}} - \phi = {{hc} \over {3{\lambda _2}}} - \phi $$ ..... (i)
and $${K_2} = {{hc} \over {{\lambda _2}}} - \phi $$ ..... (ii)
from (i) and (ii) we can say
$$3{K_1} = {K_2} - 2\phi $$
$${K_1} < {{{K_2}} \over 3}$$
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