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(B) The Einstein's photoelectric equation is given by: $K E_{\max} = \frac{hc}{\lambda} - W$,where $W$ is the work function and $\lambda$ is the wavel

Vedclass · Accepted Answer

$2 	ext{ eV}, 6200 \mathring{A}$

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(B) The Einstein's photoelectric equation is given by: $K E_{\max} = \frac{hc}{\lambda} - W$,where $W$ is the work function and $\lambda$ is the wavelength of incident light.Given $\lambda = 1240 \mathring{A} = 1240 	imes 10^{-10} 	ext{ m}$ and stopping potential $V_0 = 8 	ext{ V}$.The energy of incident photon is $E = \frac{hc}{\lambda} = \frac{12400 	ext{ eV} \cdot \mathring{A}}{1240 \mathring{A}} = 10 	ext{ eV}$.Since $K E_{\max} = eV_0 = 8 	ext{ eV}$,we have $8 	ext{ eV} = 10 	ext{ eV} - W$.Therefore,the work function $W = 10 	ext{ eV} - 8 	ext{ eV} = 2 	ext{ eV}$.The threshold wavelength $\lambda_0$ is given by $W = \frac{hc}{\lambda_0}$.$\lambda_0 = \frac{12400 	ext{ eV} \cdot \mathring{A}}{W} = \frac{12400 	ext{ eV} \cdot \mathring{A}}{2 	ext{ eV}} = 6200 \mathring{A}$.

If a photocell is illuminated with a radiation of $1240 \mathring{A}$,the stopping potential is found to be $8 \text{ V}$. Then the work function of the emitter and the threshold wavelength are:

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