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Question

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$

Vedclass · Accepted Answer

$0.61$

Vedclass · Answer

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.Let $\lambda_1 = 500 \, nm$ and $\lambda_2 = 200 \, nm$.Let $K_1$ and $K_2$ be the maximum kinetic energies corresponding to $\lambda_1$ and $\lambda_2$ respectively.Given $K_2 = 3K_1$.Using $hc = 1240 \, eV \cdot nm$:$K_1 = \frac{1240}{500} - \phi = 2.48 - \phi$$K_2 = \frac{1240}{200} - \phi = 6.2 - \phi$Substituting into $K_2 = 3K_1$:$6.2 - \phi = 3(2.48 - \phi)$$6.2 - \phi = 7.44 - 3\phi$$2\phi = 7.44 - 6.2$$2\phi = 1.24$$\phi = 0.62 \, eV$.The closest value is $0.61 \, eV$.

When the wavelength of radiation falling on a metal is changed from $500 \, nm$ to $200 \, nm$,the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to $..... \, eV$.

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