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(B) Given: Frequency of incident light $
u = 8 	imes 10^{14} \ Hz$,Maximum speed $v_{max} = 7 	imes 10^5 \ m/s$.The maximum kinetic energy of the e

Vedclass · Accepted Answer

$4.64 	imes 10^{14} \ Hz$

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(B) Given: Frequency of incident light $
u = 8 	imes 10^{14} \ Hz$,Maximum speed $v_{max} = 7 	imes 10^5 \ m/s$.The maximum kinetic energy of the emitted photoelectrons is given by $K_{max} = \frac{1}{2} m v_{max}^2$.Substituting the mass of an electron $m = 9.1 	imes 10^{-31} \ kg$:$K_{max} = \frac{1}{2} 	imes (9.1 	imes 10^{-31}) 	imes (7 	imes 10^5)^2 = 0.5 	imes 9.1 	imes 10^{-31} 	imes 49 	imes 10^{10} = 222.95 	imes 10^{-21} \ J \approx 2.23 	imes 10^{-19} \ J$.Using Einstein's photoelectric equation: $K_{max} = h
u - h
u_0$,where $
u_0$ is the threshold frequency.$h
u_0 = h
u - K_{max}$.$h
u = (6.63 	imes 10^{-34}) 	imes (8 	imes 10^{14}) = 53.04 	imes 10^{-20} \ J = 5.304 	imes 10^{-19} \ J$.$h
u_0 = 5.304 	imes 10^{-19} - 2.23 	imes 10^{-19} = 3.074 	imes 10^{-19} \ J$.Now,$
u_0 = \frac{3.074 	imes 10^{-19}}{6.63 	imes 10^{-34}} \approx 0.4636 	imes 10^{15} \ Hz = 4.64 	imes 10^{14} \ Hz$.

When light of frequency $8 \times 10^{14} \ Hz$ is incident on a surface,photoelectrons are emitted with a maximum speed of $7 \times 10^5 \ m/s$. The threshold frequency for this surface is ..........

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