(D) $1$. Saturation current $(i_s)$ is directly proportional to the intensity of incident light. Since the intensity is doubled, the saturation curren

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$i_s = 10\,mA$ and $V_s > 20\,V$

(D) $1$. Saturation current $(i_s)$ is directly proportional to the intensity of incident light. Since the intensity is doubled, the saturation current becomes $2 \times 5\,mA = 10\,mA$.$2$. Stopping potential $(V_s)$ is given by Einstein's photoelectric equation: $eV_s = h\nu - \Phi$, where $\Phi$ is the work function.$3$. Initially, $eV_{s1} = h\nu - \Phi = 10\,eV$.$4$. When frequency is doubled $(2\nu)$, the new stopping potential $V_{s2}$ satisfies $eV_{s2} = h(2\nu) - \Phi = 2h\nu - \Phi$.$5$. Since $h\nu = 10 + \Phi$, we have $eV_{s2} = 2(10 + \Phi) - \Phi = 20 + 2\Phi - \Phi = 20 + \Phi$.$6$. Since $\Phi > 0$, it follows that $V_{s2} > 20\,V$.$7$. Therefore, $i_s = 10\,mA$ and $V_s > 20\,V$.

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

Difficult

In the experiment of $P.E.E.$ (Photoelectric Effect), the saturation current is $5\,mA$ and the stopping potential is $10\,V$. If the intensity and frequency of the incident light are both doubled, then what will be the new saturation current $(i_s)$ and stopping potential $(V_s)$?

A
$i_s = 5\,mA$ and $V_s = 10\,V$
B
$i_s = 10\,mA$ and $V_s = 20\,V$
C
$i_s = 5\,mA$ and $V_s > 20\,V$
D
$i_s = 10\,mA$ and $V_s > 20\,V$

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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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In the experiment of $P.E.E.$ (Photoelectric Effect), the saturation current is $5\,mA$ and the stopping potential is $10\,V$. If the intensity and frequency of the incident light are both doubled, then what will be the new saturation current $(i_s)$ and stopping potential $(V_s)$?

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$A$ photosensitive metallic surface has a work function $h\nu_0$. If photons of energy $2h\nu_0$ fall on this surface,the electrons are emitted with a maximum velocity of $4 \times 10^6 \, m/s$. When the photon energy is increased to $5h\nu_0$,the maximum velocity of the photoelectrons will be:

In the photoelectric effect experiment, if the frequency of incident radiation $(v)$ is increased, keeping all other factors constant, what happens to the stopping potential $(V_s)$, given $(v > v_0)$ where $(v_0)$ is the threshold frequency?

The energy of a photon with wavelength $\lambda$ is $2 \ eV$. When it strikes a metal surface,the maximum velocity of the emitted photoelectrons is $v$. If the value of $\lambda$ is decreased by $25\%$ and the maximum velocity is doubled,the work function of the metal becomes ...... $eV$.

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