(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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Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with kinetic energy $K$. If $K$ is to be increased to twice its original value,the wavelength must be changed to $\lambda'$ such that :-

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When a photon of energy $8 \ eV$ is incident on a metal surface with a threshold frequency of $1.6 \times 10^{15} \ Hz$,the maximum kinetic energy of the emitted photoelectrons is .......... $eV$. (Given: $h = 6.6 \times 10^{-34} \ Js, 1 \ eV = 1.6 \times 10^{-19} \ J$)

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The slope of the graph between stopping potential and the frequency of incident radiation for the photoelectric effect is . . . . . . .

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Light of two different frequencies whose photons have energies $1.3 eV$ and $2.8 eV$ respectively,successfully illuminate a metallic surface whose work function is $0.8 eV$. The ratio of maximum speeds of emitted electrons will be

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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 $\lambda$ strikes a photoelectric surface and electrons are ejected with kinetic energy $K$. If $K$ is to be increased to twice its original value,the wavelength must be changed to $\lambda'$ such that :-

Light of wavelength $\lambda$ is incident on the surface of a metal having work function $\phi$, causing the emission of electrons. What is the maximum velocity of the emitted electrons? (Given: $c = \text{velocity of light}$, $h = \text{Planck's constant}$, $m = \text{mass of electron}$)

When a photon of energy $8 \ eV$ is incident on a metal surface with a threshold frequency of $1.6 \times 10^{15} \ Hz$,the maximum kinetic energy of the emitted photoelectrons is .......... $eV$. (Given: $h = 6.6 \times 10^{-34} \ Js, 1 \ eV = 1.6 \times 10^{-19} \ J$)

The slope of the graph between stopping potential and the frequency of incident radiation for the photoelectric effect is . . . . . . .

Light of two different frequencies whose photons have energies $1.3 eV$ and $2.8 eV$ respectively,successfully illuminate a metallic surface whose work function is $0.8 eV$. The ratio of maximum speeds of emitted electrons will be

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