(B) Given, work function $\phi = 2.4 \text{ eV}$.Energy required to ionize a hydrogen atom in the ground state is $E_i = 13.6 \text{ eV}$.For the hydr

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(B) Given, work function $\phi = 2.4 \text{ eV}$.Energy required to ionize a hydrogen atom in the ground state is $E_i = 13.6 \text{ eV}$.For the hydrogen atom to be ionized by the photoelectron, the kinetic energy $(KE)$ of the photoelectron must be at least equal to the ionization energy of the hydrogen atom.Thus, $KE_{max} = 13.6 \text{ eV}$.According to the photoelectric equation, $E = KE_{max} + \phi$, where $E = \frac{hc}{\lambda}$.Substituting the values: $\frac{1240}{\lambda} = 13.6 + 2.4$.$\frac{1240}{\lambda} = 16$.$\lambda = \frac{1240}{16} = 77.5 \text{ nm}$.

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

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$A$ monochromatic light of wavelength $\lambda$ ejects photoelectrons from a metal surface with work function $\phi = 2.4 \text{ eV}$. These photoelectrons are made to collide with hydrogen atoms in the ground state. The maximum value of $\lambda$ for which the hydrogen atom may be ionized is [take $hc = 1240 \text{ eV-nm}$]. (in $\text{ nm}$)

TS EAMCET 2020 All TS EAMCET

A
$80$
B
$77.5$
C
$75.5$
D
$85$

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Class 12

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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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The work function of nickel is $5 \text{ eV}$. When light of wavelength $2000 \text{ Å}$ falls on it,it emits photoelectrons. The potential difference necessary to stop the fastest emitted electrons is (given $h = 6.67 \times 10^{-34} \text{ J-s}$): (in $\text{ V}$)

DifficultAP EAMCET 2007

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In an experiment on the photoelectric effect,the frequency $\nu$ of the incident light is plotted against the stopping potential $V_0$. The work function of the photoelectric surface is given by ($e$ is the electronic charge):

Easy

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When light of frequency twice the threshold frequency is incident on a metal plate,the maximum velocity of the emitted electron is $v_{1}$. When the frequency of incident radiation is increased to five times the threshold value,the maximum velocity of the emitted electron becomes $v_{2}$. If $v_{2} = x v_{1}$,the value of $x$ will be...........

MediumJEE MAIN 2022

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Two identical photocathodes receive light of frequencies $f_{1}$ and $f_{2}$ respectively. If the velocities of the photo-electrons emitted are $v_{1}$ and $v_{2}$ respectively,then:

Difficult

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The threshold wavelength for the photoelectric effect of a metal is $6500 \mathring{A}$. The work function of the metal is approximately .......... $eV$.

Easy

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The work function of nickel is $5 \text{ eV}$. When light of wavelength $2000 \text{ Å}$ falls on it,it emits photoelectrons. The potential difference necessary to stop the fastest emitted electrons is (given $h = 6.67 \times 10^{-34} \text{ J-s}$): (in $\text{ V}$)

In an experiment on the photoelectric effect,the frequency $\nu$ of the incident light is plotted against the stopping potential $V_0$. The work function of the photoelectric surface is given by ($e$ is the electronic charge):

Two identical photocathodes receive light of frequencies $f_{1}$ and $f_{2}$ respectively. If the velocities of the photo-electrons emitted are $v_{1}$ and $v_{2}$ respectively,then:

The threshold wavelength for the photoelectric effect of a metal is $6500 \mathring{A}$. The work function of the metal is approximately .......... $eV$.

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