Class 12PhysicsDual Nature of Radiation and matterEinstein's Photoelectric Equation and Energy Quantum of Radiation
DifficultLight source having wavelength $331 \text{ nm}$ is used to generate photo-electrons whose stopping potential is $0.2 \text{ V}$. The work function of the used metal in the experiment is $\alpha \times 10^{-19} \text{ J}$. The value of $\alpha$ is . . . . . . . ($h = 6.62 \times 10^{-34} \text{ J s}$, $e = 1.6 \times 10^{-19} \text{ C}$ and $c = 3 \times 10^8 \text{ m/s}$) (in $.68$)
- A$3$
- B$4$
- C$5$
- D$2$
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Similar Questions
The work function of a certain metal is $3.31 \times 10^{-19} \,J$. The maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 \text{ Å}$ is (Given: $h = 6.62 \times 10^{-34} \,J \cdot s$,$c = 3 \times 10^8 \,m/s$,$e = 1.6 \times 10^{-19} \,C$) (in $\text{ eV}$)
DifficultAP EAMCET 2009
View SolutionRadiation of wavelength $332 \ nm$ is incident on a metal surface having a work function of $1.07 \ eV$. The stopping potential required to stop the emission of photoelectrons from the metal surface is ............ $V$. $(h = 6.6 \times 10^{-34} \ J s, c = 3 \times 10^8 \ m/s, 1 \ eV = 1.6 \times 10^{-19} \ J)$
Difficult
View SolutionWhen a certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential for photoelectric current is $4 V_0$. When the same surface is illuminated with light of wavelength $3 \lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is
Light of wavelength $5000 \ \mathring A$ and intensity $4.68 \ mW/cm^2$ is incident on a photosensitive surface. If only $5\%$ of the incident photons produce photoelectrons,what is the number of photoelectrons emitted per unit area per unit time?
Difficult
View SolutionLet $v_1$ and $v_2$ be the maximum velocities of the emitted electrons when the surface of a metal is illuminated with light waves of energy $E_1 = 4 \text{ eV}$ and $E_2 = 2.5 \text{ eV}$,respectively. If the work function of the metal is $2 \text{ eV}$,then the ratio $\frac{v_1}{v_2}$ is
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