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(C) Given:Threshold frequency $
u_{0} = 3.3 	imes 10^{14} \;Hz$Incident frequency $
u = 8.2 	imes 10^{14} \;Hz$According to Einstein's photoelectr

Vedclass · Accepted Answer

$2.03$

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(C) Given:Threshold frequency $
u_{0} = 3.3 	imes 10^{14} \;Hz$Incident frequency $
u = 8.2 	imes 10^{14} \;Hz$According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = h(
u - 
u_{0})$.The cutoff voltage (stopping potential) $V_{0}$ in $eV$ is numerically equal to the maximum kinetic energy in $eV$,where $eV_{0} = h(
u - 
u_{0})$.Using Planck's constant $h \approx 4.135 	imes 10^{-15} \;eV \cdot s$:$V_{0} = \frac{h(
u - 
u_{0})}{e} = 4.135 	imes 10^{-15} 	imes (8.2 - 3.3) 	imes 10^{14}$$V_{0} = 4.135 	imes 10^{-15} 	imes 4.9 	imes 10^{14}$$V_{0} = 4.135 	imes 0.49 \approx 2.026 \;V$.Rounding to two decimal places,the cutoff voltage is $2.03 \;V$.

The threshold frequency for a certain metal is $3.3 \times 10^{14} \;Hz$. If light of frequency $8.2 \times 10^{14} \;Hz$ is incident on the metal,predict the cutoff voltage (in $eV$) for the photoelectric emission.

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