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(B) The work function $\Phi$ is given by the formula $\Phi = \frac{hc}{\lambda_0}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\l

Vedclass · Accepted Answer

$1.98$

Vedclass · Answer

(B) The work function $\Phi$ is given by the formula $\Phi = \frac{hc}{\lambda_0}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda_0$ is the threshold wavelength.Given: $h = 6.6 	imes 10^{-34} \ 	ext{Js}$, $c = 3 	imes 10^8 \ 	ext{m/s}$, and $\lambda_0 = 6250 \ 	ext{Å} = 6250 	imes 10^{-10} \ 	ext{m}$.Substituting the values:$\Phi = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{6250 	imes 10^{-10}} \ 	ext{J}$.$\Phi = \frac{19.8 	imes 10^{-26}}{6250 	imes 10^{-10}} = \frac{19.8}{6250} 	imes 10^{-16} \ 	ext{J} = 3.168 	imes 10^{-20} \ 	ext{J}$.To convert this into electron-volts (eV), divide by the charge of an electron $e = 1.6 	imes 10^{-19} \ 	ext{C}$:$\Phi = \frac{3.168 	imes 10^{-20}}{1.6 	imes 10^{-19}} \ 	ext{eV} = 1.98 \ 	ext{eV}$.Thus, the correct option is $B$.

If the threshold wavelength of light for photoelectric emission to take place from a metal surface is $6250 \ \text{Å}$, then the work function of the metal is (Planck's constant $= 6.6 \times 10^{-34} \ \text{Js}$) (in $\text{eV}$)

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