Dual Nature Of Radiation - Modern Physics - Physics Previous Year Questions

A metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value then the maximum KE of the emitted photoelectrons would be

When photons of energy $h v$ fall on a metal plate of work function ' $W_0$ ', photoelectrons of maximum kinetic energy ' $K$ ' are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be

The maximum velocity of the photoelectron emitted by the metal surface is ' $v$ '. Charge and mass of the photoelectron is denoted by ' $e$ ' and ' $m$ ' respectively. The stopping potential in volt is

The stopping potential of the photoelectrons, from a photo cell is

When certain metal surface is illuminated with a light of wavelength $\lambda$, the stopping potential is $V$, When the same surface is illuminated by light of wavelength $2 \lambda$, the stopping potential is $\left(\frac{V}{3}\right)$. The threshold wavelength for the surface is

The graph of kinetic energy against the frequency \(v\) of incident light is as shown in the figure. The slope of the graph and intercept on \(X\)-axis respectively are

The light of wavelength \(\lambda\) incident on the surface of metal having work function \(\phi\) emits the electrons. The maximum velocity of electrons emitted is [ \(c=\) velocity of light, \(h=\) Planck's constant, \(m=\) mass of electron]

Energy of the incident photon on the metal surface is \(3 W\) and then \(5 W\), where \(W\) is the work function for that metal. The ratio of velocities of emitted photoelectrons is

The maximum velocity of the photoelectron emitted by the metal surface is \(v\). Charge and mass of the photoelectron is denoted by \(e\) and \(m\), respectively. The stopping potential in volt is

The graph of stopping potential $V_s$ against frequency $v$ of incident radiation is plotted for two different metals $P$ and $Q$ as shown in the graph. $\phi_p$ and $\phi_Q$ are work-functions of $P$ and $Q$ respectively, then

If the maximum kinetic energy of emitted electrons in photoelectric effect is $3.2 \times 10^{-19} \mathrm{~J}$ and the work-function for metal is $6.63 \times 10^{-19} \mathrm{~J}$, then stopping potential and threshold wavelength respectively are

[Planck's constant, $h=6.63 \times 10^{34} \mathrm{~J}$-s]

[Velocity of light, $c=3 \times 10^8 \frac{\mathrm{~m}}{\mathrm{~s}}$ ]

[Charge on electron $=1.6 \times 10^{-19} \mathrm{C}$ ]

What is the additional energy that should be supplied to a moving electron to reduce its de Broglie wavelength from \(1 \mathrm{~nm}\) to \(0.5 \mathrm{~nm}\) ?

Photoelectrons are emitted when photons of energy \(4.2 ~\mathrm{eV}\) are incident on a photosensitive metallic sphere of radius \(10 \mathrm{~cm}\) and work function \(2.4 ~\mathrm{eV}\). The number of photoelectrons emitted before the emission is stopped is

\(\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9\right.\) SI unit; \(\left.\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right]\)

When a photosensitive surface is irradiated by light of wavelengths '\(\lambda_1\)' and '\(\lambda_2\)', kinetic energies of emitted photoelectrons are 'E\(_1\)' and 'E\(_2\)' respectively. The work function of photosensitive surface is

When light of wavelength '\(\lambda\)' is incident on a photosensitive surface, photons of power 'P' are emitted. The number of photon 'n' emitted in time 't' is [h = Planck's constant, c = velocity of light in vacuum]

Photoemission from metal surface takes place for frequencies '\(v_1\)' and '\(v_2\)' of incident rays \(\left(v_1>v_2\right)\). Maximum kinetic energy of photoelectrons emitted is in the ratio \(1: \mathrm{K}\). The threshold frequency of metallic surface is

The wave number of the last line of the Balmer series in the hydrogen spectrum will be \(\left(\right.\) Rydberg's cons \(\left.\tan t, R=\frac{10^7}{\mathrm{~m}}\right)\)

A proton and alpha particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength of proton to that of alpha particle will be (mass of alpha particle is four times mass of proton.)

Light of frequency two times the threshold frequency is incident on photosensitive material. If the incident frequency is made \(\left(\frac{1}{3}\right)^{\text {rd }}\) and intensity is doubled, then the photoelectric current will

On a photosensitive surface, if the intensity of incident radiation is increased, the stopping potential