When a metal plate is illuminated with light | vedclass

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(A) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = W_0 + K_{max}$,where $K_{max} = \frac{1}{2}mv^2$.For $\lambda_1 = 400 \ nm =

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$2 hc 	imes 10^6 \ J$

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(A) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = W_0 + K_{max}$,where $K_{max} = \frac{1}{2}mv^2$.For $\lambda_1 = 400 \ nm = 400 	imes 10^{-9} \ m$ and velocity $v_1 = v$:$\frac{hc}{400 	imes 10^{-9}} = W_0 + \frac{1}{2}mv^2 \quad \dots(i)$For $\lambda_2 = 250 \ nm = 250 	imes 10^{-9} \ m$ and velocity $v_2 = 2v$:$\frac{hc}{250 	imes 10^{-9}} = W_0 + \frac{1}{2}m(2v)^2 = W_0 + 2mv^2 \quad \dots(ii)$From $(i)$,$\frac{1}{2}mv^2 = \frac{hc}{400 	imes 10^{-9}} - W_0$. Substituting this into $(ii)$:$\frac{hc}{250 	imes 10^{-9}} = W_0 + 4 \left( \frac{hc}{400 	imes 10^{-9}} - W_0 ight)$$\frac{hc}{250 	imes 10^{-9}} = W_0 + \frac{hc}{100 	imes 10^{-9}} - 4W_0$$3W_0 = hc \left( \frac{1}{100 	imes 10^{-9}} - \frac{1}{250 	imes 10^{-9}} ight)$$3W_0 = hc \left( \frac{2.5 - 1}{250 	imes 10^{-9}} ight) = hc \left( \frac{1.5}{250 	imes 10^{-9}} ight) = hc \left( \frac{15}{2500 	imes 10^{-9}} ight) = hc \left( \frac{3}{500 	imes 10^{-9}} ight)$$W_0 = \frac{hc}{500 	imes 10^{-9}} = 0.2 	imes 10^7 \ hc = 2 	imes 10^6 \ hc \ J$.

When a metal plate is illuminated with light of wavelengths $400 \ nm$ and $250 \ nm$,the maximum velocities of the emitted photoelectrons are $v$ and $2v$,respectively. The work function of the metal is ($h$ = Planck's constant; $c$ = speed of light in vacuum):

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