Dual nature of radiation and matter practice · Medium

Assertion: Through light of a single frequency (mono-chromatic) is incident on a metal, the energies of emitted photoelectrons are different.

Reason: the energy of electrons emitted from inside the metal surface is lost in collision with the other atoms in the metal.

When light falls on a metal surface, the maximum kinetic energy of the emitted photo-electrons depends upon

Assertion: The wave nature of electrons was first experimentally verified by Davisson and Germer Experiment.

Reason: From the electron diffraction measurements, the wavelength of matter waves was found to be $0.165 \mathrm{~nm}$.

A particle which has zero rest mass and non-zero energy and momentum, must travel with a speed

For the Bohr's first orbit of circumference $2\pi r,$ the de-Broglie wavelength of revolving electron will be.

A photon is emitted by a hydrogen atom when it comes from excited state $\mathrm{n}=5$ to the ground state. The recoil speed is almost

Which of the following is not possible value of radius of stable orbit of hydrogen atom. Where $\lambda$ is de Broglie wavelength of the electron.

Statement 1: Mass of moving photon varies inversely as the wavelength.

Statement 2: Energy of the particle $=$ Mass $\times(\text { Speed of light })^2$

In photoelectric effect, the KE of electrons emitted from the metal surface depends upon

Photoelectric effect supports quantum nature of light because
1. There is minimum frequency of light below which no photoelectrons are emitted.
2. Electric charge of photoelectrons is quantized.
3. Maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity.
4. Even when metal surface is faintly illuminated the photoelectrons leave the surface immediately.

An electron with mass 'm' with an initial velocity $(\text{t}=0)$ $\vec{\mathrm{v}}=\mathrm{v}_0 \hat{\mathrm{i}} \left(\mathrm{v}_0>0\right)$ enters a magnetic field $\vec{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{j}}.$ If the initial de-Broglie wavelength at $\mathrm{t}=0$ is $\lambda_0$ then its value after time 't' would be:

Ultraviolet radiation of 6.2 eV falls on an aluminium surface (work function 4.2 eV). The kinetic energy of the faster electron emitted is approximately

The number of photo-electrons emitted per second from a metal surface increases when

In cathode ray oscillograph, the focusing of beam on the screen is achieved by

When ultraviolet rays are incident on metal plate, then photoelectric effect does not occur. It occurs by the incidence of

If a photoemissive surface has a threshold frequency of $4.6 \times 10^{14} \mathrm{~Hz}$, calculate the energy of the photons in $\mathrm{eV}$. Given $h=6.6 \times 10^{-34} \mathrm{~Js}$.

Vidicon works on the principle of

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength $491nm\,$ is $0.710V$. When the incident wavelength is changed to a new value, the stopping potential is $1.43V.$ The new wavelength is

Which is the correct ascending order of wavelengths?

A particle with rest mass zero is moving speed $c$. The de-Broglie wavelength associated with it

An $\alpha \mathrm{ - Particle}$ and a deuteron are moving with velocities v and 2v respectively. What will be the ratio of their de Broglie wavelengths-?

A metal plate gets heated when cathode rays strike against it due to

The energy of a photon is $E = hv$ and the momentum of photon $p = \frac{h}{\lambda },$ then the velocity of photon will be

Millikan’s oil-drop experiment established that

The difference between kinetic energies of photoelectrons emitted from a surface by light of wavelength 2500$\mathop A\limits^ \circ$ and 5000$\mathop A\limits^ \circ$ will be

Which one of the following statements regarding photo-emission of electrons is correct?

Which of the following is not the property of a cathode ray

Assertion: The momentum of a photon is equal to $\mathrm{p}=\mathrm{h} / \lambda$.

Reason: According to Planck $\mathrm{E}=\mathrm{hv}$

According to Einstein $\mathrm{E}=\mathrm{mc}^2$.

So that $\mathrm{mc}^2=\mathrm{hv} \Rightarrow \mathrm{p}=\mathrm{mc}=\mathrm{h} / \lambda$

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $\lambda_1$ and $\lambda_2$ their de Broglie wavelength in the frame of reference attached to their centre of mass is :

The slope of frequency of incident light and stopping potential graph for a given surface will be