A photoelectric material having work-function | vedclass

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Question

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the photoelectron is given by:$K_{max} = \frac{hc}{\lambda}

Vedclass · Accepted Answer

$\lambda_d^3 / \lambda^2$

Vedclass · Answer

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the photoelectron is given by:$K_{max} = \frac{hc}{\lambda} - \phi_0$Since $K_{max} = \frac{p^2}{2m}$ and the de Broglie wavelength is $\lambda_d = \frac{h}{p}$,we have $p = \frac{h}{\lambda_d}$.Substituting this into the kinetic energy equation:$\frac{h^2}{2m \lambda_d^2} = \frac{hc}{\lambda} - \phi_0$Differentiating both sides with respect to $\lambda$ and $\lambda_d$:$\frac{d}{d\lambda_d} \left( \frac{h^2}{2m \lambda_d^2} \right) d\lambda_d = \frac{d}{d\lambda} \left( \frac{hc}{\lambda} - \phi_0 \right) d\lambda$$\frac{h^2}{2m} (-2 \lambda_d^{-3}) d\lambda_d = -hc \lambda^{-2} d\lambda$$\frac{h^2}{m \lambda_d^3} d\lambda_d = \frac{hc}{\lambda^2} d\lambda$Rearranging for the ratio $\frac{d\lambda_d}{d\lambda}$:$\frac{d\lambda_d}{d\lambda} = \frac{hc}{\lambda^2} \cdot \frac{m \lambda_d^3}{h^2} = \left( \frac{mc}{h} \right) \frac{\lambda_d^3}{\lambda^2}$Thus,$\frac{\Delta \lambda_d}{\Delta \lambda} \propto \frac{\lambda_d^3}{\lambda^2}$.

Similar Questions

The work function of metals is in the range of $2 eV$ to $5 eV$. Find which of the following wavelengths of light cannot be used for the photoelectric effect (in $nm$)? (Consider, Planck constant $= 4 \times 10^{-15} eVs$, velocity of light $= 3 \times 10^{8} m/s$)

Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with energy $E$. If $E$ is to be increased to twice the original value,the wavelength changes to $\lambda_1$. Which of the following is correct?

Let $K_1$ be the maximum kinetic energy of photoelectrons emitted by a light of wavelength $\lambda_1$ and $K_2$ corresponding to $\lambda_2$. If $\lambda_1 = 2\lambda_2$,then:

The graph of stopping potential $(V_{s})$ against frequency $(\nu)$ of incident radiation is plotted for two different metals '$P$' and '$Q$' as shown in the graph. If $\phi_{P}$ and $\phi_{Q}$ are the work functions of metals '$P$' and '$Q$' respectively,then which of the following is correct?

The photoelectric threshold wavelength for a metal surface is $6600 \, \mathring{A}$. The work function for this is:

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