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(B) The energy of photons corresponding to wavelength $\lambda_1 = 2475 \ \mathring{A}$ is $E_1 = \frac{12375}{2475} = 5 \ eV$. The energy of photons

Vedclass · Accepted Answer

$5$

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(B) The energy of photons corresponding to wavelength $\lambda_1 = 2475 \ \mathring{A}$ is $E_1 = \frac{12375}{2475} = 5 \ eV$. The energy of photons corresponding to wavelength $\lambda_2 = 6000 \ \mathring{A}$ is $E_2 = \frac{12375}{6000} = 2.06 \ eV$. Since $E_2  W_0$,photoelectric emission is possible only with $\lambda_1$. The maximum kinetic energy of the emitted photoelectrons is $K = E_1 - W_0 = 5 - 4.8 = 0.2 \ eV$. Converting $K$ to Joules: $K = 0.2 	imes 1.6 	imes 10^{-19} \ J = 3.2 	imes 10^{-20} \ J$. The radius of the circular path in a magnetic field is given by $r = \frac{\sqrt{2mK}}{qB}$. Substituting the values: $r = \frac{\sqrt{2 	imes 9 	imes 10^{-31} 	imes 3.2 	imes 10^{-20}}}{1.6 	imes 10^{-19} 	imes 3 	imes 10^{-5}}$. $r = \frac{\sqrt{57.6 	imes 10^{-51}}}{4.8 	imes 10^{-24}} = \frac{7.59 	imes 10^{-25}}{4.8 	imes 10^{-24}} \approx 0.05 \ m = 5 \ cm$.

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