(A) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K.E_{\max} = E - \phi$. For the first case,$E_1 = 3\phi$,so

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K.E_{\max} = E - \phi$. For the first case,$E_1 = 3\phi$,so $K.E_1 = 3\phi - \phi = 2\phi$. For the second case,$E_2 = 9\phi$,so $K.E_2 = 9\phi - \phi = 8\phi$. Since $K.E = \frac{1}{2}mv^2$,we have the ratio $\frac{K.E_1}{K.E_2} = \frac{v_1^2}{v_2^2}$. Substituting the values,$\frac{2\phi}{8\phi} = \frac{1}{4} = \frac{v_1^2}{v_2^2}$. Taking the square root,$\frac{v_1}{v_2} = \frac{1}{2}$,which implies $v_2 = 2v_1$. Given $v_1 = 6 \times 10^6 \ m/s$,then $v_2 = 2 \times 6 \times 10^6 = 12 \times 10^6 \ m/s$.

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

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$A$ photosensitive metallic surface has a work function $\phi$. If a photon of energy $3 \phi$ falls on the surface,the electron is emitted with a maximum velocity of $6 \times 10^6 \ m/s$. When the photon energy is increased to $9 \phi$,the maximum velocity of the photoelectrons will be:

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A
$12 \times 10^6 \ m/s$
B
$6 \times 10^6 \ m/s$
C
$3 \times 10^6 \ m/s$
D
$24 \times 10^6 \ m/s$

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

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$A$ photosensitive metallic surface has a work function $\phi$. If a photon of energy $3 \phi$ falls on the surface,the electron is emitted with a maximum velocity of $6 \times 10^6 \ m/s$. When the photon energy is increased to $9 \phi$,the maximum velocity of the photoelectrons will be:

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Light of wavelength $4000 \ \mathring A$ is incident on a photosensitive surface. If a potential of $-2 \ V$ is required to stop the emitted electrons,the work function of the material is: $(h = 6.6 \times 10^{-34} \ J \cdot s, e = 1.6 \times 10^{-19} \ C, c = 3 \times 10^8 \ m/s)$ (in $eV$)

The stopping potential $(V_0)$ versus frequency $(\nu)$ plot of a substance is shown in the figure. The threshold wavelength is:

$A$ metal surface of work function $1.13 eV$ is irradiated with light of wavelength $310 nm$. The retarding potential required to stop the escape of photoelectrons is [Take $hc = 1240 eV \cdot nm$] (in $V$)

Two photons of energy $2.5 eV$ and $3.5 eV$ fall on a metal surface of work function $1.5 eV$. The ratio of the maximum velocities of the photoelectrons emitted from the metal surface is

When a metal surface is illuminated by light of wavelength $\lambda$,the stopping potential is $8 \ V$. When the same surface is illuminated by light of wavelength $3 \lambda$,the stopping potential is $2 \ V$. The threshold wavelength for this surface is:

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