(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $K_{max} = eV_s$ ($V_s$ is the stopping potential).For

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(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $K_{max} = eV_s$ ($V_s$ is the stopping potential).For the first case: $\frac{hc}{\lambda} = 5eV_0 + \phi$ --- $(1)$For the second case: $\frac{hc}{3\lambda} = eV_0 + \phi$ --- $(2)$Multiply equation $(2)$ by $5$: $\frac{5hc}{3\lambda} = 5eV_0 + 5\phi$ --- $(3)$Subtract equation $(1)$ from equation $(3)$:$\frac{5hc}{3\lambda} - \frac{hc}{\lambda} = (5eV_0 + 5\phi) - (5eV_0 + \phi)$$\frac{5hc - 3hc}{3\lambda} = 4\phi$$\frac{2hc}{3\lambda} = 4\phi$$\phi = \frac{2hc}{12\lambda} = \frac{hc}{6\lambda}$

Class 12 Physics Dual Nature of Radiation and matter Einstein's Photoelectric Equation and Energy Quantum of Radiation

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When a metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential is $5V_0$. When the same surface is illuminated with light of wavelength $3\lambda$,the stopping potential is $V_0$. Then the work function of the metallic surface is:

A
$\frac{hc}{6\lambda}$
B
$\frac{hc}{5\lambda}$
C
$\frac{hc}{4\lambda}$
D
$\frac{2hc}{4\lambda}$

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Class 12

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Chapter

Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

When a light of wavelength $\lambda$ falls on the emitter of a photocell,the maximum speed of emitted photoelectrons is $V$. If the incident wavelength is changed to $2\lambda / 3$,the maximum speed of the emitted photoelectrons will be:

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If a metal sheet is irradiated with radiations of frequencies $v_1$ and $v_2$ and the kinetic energies of the emitted photoelectrons are in the ratio $1 : x$, then the threshold frequency of the metal is:

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$A$ certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for the photoelectric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is:

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The maximum velocity of the photoelectron emitted by the metal surface is $v$. The charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volts is:

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The threshold wavelength for photoelectric emission from a material is $5500\,\mathring A$. Photoelectrons will be emitted when this material is illuminated with monochromatic radiation from a:
$A.$ $75\,W$ infra-red lamp
$B.$ $10\,W$ infra-red lamp
$C.$ $75\,W$ ultra-violet lamp
$D.$ $10\,W$ ultra-violet lamp
Choose the correct answer from the options given below:

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When a metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential is $5V_0$. When the same surface is illuminated with light of wavelength $3\lambda$,the stopping potential is $V_0$. Then the work function of the metallic surface is:

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When a light of wavelength $\lambda$ falls on the emitter of a photocell,the maximum speed of emitted photoelectrons is $V$. If the incident wavelength is changed to $2\lambda / 3$,the maximum speed of the emitted photoelectrons will be:

If a metal sheet is irradiated with radiations of frequencies $v_1$ and $v_2$ and the kinetic energies of the emitted photoelectrons are in the ratio $1 : x$, then the threshold frequency of the metal is:

The maximum velocity of the photoelectron emitted by the metal surface is $v$. The charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volts is:

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