Class 12PhysicsDual Nature of Radiation and MatterEinstein's Photoelectric Equation and Energy Quantum of Radiation
Medium$A$ photoelectric cell is connected to a source of variable potential difference and the resulting photoelectric current $(\mu A)$ is plotted against the applied potential difference $(V)$. The graph with the broken line represents one situation for a given frequency and intensity of the incident radiation. If the frequency is increased and intensity is reduced,which of the following graphs of unbroken line represents the new situation?
- A$A$
- B$B$
- C$C$
- D$D$
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Similar Questions
Photoelectric emission takes place from a certain metal at threshold frequency $v$. If the radiation of frequency $4v$ is incident on the metal plate,the maximum velocity of the emitted photoelectrons will be ($m=$ mass of photoelectron,$h=$ Planck's constant).
In a photoelectric experiment, the wavelength of the light incident on the metal is changed from $200 \, nm$ to $400 \, nm$. The decrease in the stopping potential is close to [Use $hc = 1240 \, eV \cdot nm$ where $h$ is Planck's constant and $c$ is the velocity of light]. (in $ \, V$)
Assuming photoemission to take place,the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times,is
Medium
View Solution$A$ beam of light of wavelength $400\,nm$ and power $1.55\,mW$ is directed at the cathode of a photoelectric cell. If only $10\%$ of the incident photons effectively produce photoelectrons,then find the current due to these electrons in $\mu A$. (Given: $hc = 1240\,eV\cdot nm$,$e = 1.6 \times 10^{-19}\,C$)
MediumAIIMS 2015
View SolutionThe threshold frequency of a metal is $f_0$. When light of frequency $2f_0$ is incident on the metal plate,the maximum velocity of the photoelectrons is $v_1$. When the frequency of the incident radiation is increased to $5f_0$,the maximum velocity of the photoelectrons emitted is $v_2$. The ratio of $v_1$ to $v_2$ is:
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