$\lambda_0=\frac{\text{hc}}{\text{W}_0}$
Where,
$\lambda_0$ = threshold wavelength
h = Planck’s constant
c = velocity of radiation
Substituting the values in the given expression of $(\lambda_0):$
$\lambda_0=\frac{(6.626\times10^{-34}\text{Js})(3.0\times10^8\text{ms}^{-1})}{1.9\times1.602\times10^{-19}\text{J}}$
$\lambda_0=6.53\times10^{-1}\text{m}$
Hence, the threshold wavelength $\lambda_0$ is 653nm.
$\text{v}_0=\frac{\text{W}_0}{\text{h}}$
Where,
ν0 = threshold frequency
h = Planck’s constant
Substituting the values in the given expression of ν0:
$\text{v}_0=\frac{1.9\times1.602\times10^{-19}\text{J}}{6.626\times10^{-34}\text{Js}}$
(1eV = 1.602 × 10–19J)
ν0 = 4.593 × 1014s–1
Hence, the threshold frequency of radiation (ν0) is 4.593 × 1014s–1.
According to the question: Wavelength used in irradiation (λ) = 500nm Kinetic energy = h (ν – ν0) $=\text{hc}\Big(\frac{1}{\lambda}-\frac{1}{\lambda_0}\Big)$ $=(6.626\times10^{-34}\text{Js})(3.0\times10^8\text{ms}^{-1})\Big(\frac{\lambda_0-\lambda}{\lambda\lambda_0}\Big)$ $=(1.9878\times10^{-26}\text{Js})\Big[\frac{(653-500)10^{-9}\text{m}}{(653)(500)10^{-18}\text{m}^2}\Big]$ $=\frac{(1.9878\times10^{-26})(153\times10^{9})}{(653)(500)}\text{J}$ $=9.3149\times10^{-20}\text{J}$ Kinetic energy of the ejected photoelectron $=9.3149\times10^{-20}\text{J}$ Since $\text{K.E}=\frac{1}{2}\text{mv}^2=9.3149\times10^{-20}\text{J}$ $\text{v}=\sqrt{\frac{2(9.3149\times10^{-20}\text{J})}{9.10939\times10^{-31}\text{kg}}}$ $=\sqrt{2.0451\times10^{11}\text{m}^2\text{s}^{-2}}$ $\text{v}=4.52\times10^5\text{ms}^{-1}$ Hence, the velocity of the ejected photoelectron (v) is $\text{v}=4.52\times10^5\text{ms}^{-1}.$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.