Dual nature of radiation and matter practice · Mixed difficulty

When a photon of wavelength 400 nm is incident on a metal surface, the electron emission takes place. The stopping potential to stop electron is 1.5 V. The work function of metal surface is: (given $h c=1240 \mathrm{~eV}-\mathrm{nm})$

Assertion: An electron cannot be observed without changing its momentum by an indeterminate amount.

Reason: Momentum of an electron is $\frac{h}{\lambda}$.

Kinetic mass of photon is

If the momentum of an electron is changed by $\Delta p,$ then the de-Broglie wavelength associated with it changes by 0.50%. The initial momentum of the electron will be

If $c$ is the velocity of light in free space, the correct statements about photon among the following are:

A. The energy of a photon is $E=h v$

B. The velocity of a photon is $c$

C. The momentum of a photon, $p=\frac{h v}{c}$

D. In a photon-electron collision, both total energy and total momentum are conserved.

E. Photon possesses positive charge.

Choose the correct answer from the options given below:

In Millikans oil drop experiment, a charged drop of mass $1.8 \times {10^{ - 14}}kg$ is stationary between its plates. The distance between its plates is $0.90\;cm$ and potential difference is $2.0\;kilo\;volts.$ The number of electrons on the drop is

The increase in the energy of an electron for changing its de-Broglie wavelength from $1\,\mathop \mathrm{A}\limits^\mathrm{o} \mathrm{ }\,\,\mathrm{to}\,\,\mathrm{ }0.5\,\mathop \mathrm{A}\limits^\mathrm{o} \,$ will be-

A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of $\lambda$. An alpha particle having certain kinetic energy has the same de-Brogle wavelength $\lambda$. The ratio of kinetic energy of proton and that of alpha particle is

According to photon theory of light which of the following physical quantities associated with a photon do not/does not change as it collides with an electron in vacuum

Assertion: Photoelectric effect supports the quantum nature of light.

Reason: Photoelectric emission is instantaneous.

When a metallic surface is illuminated with monochromatic light of wavelength $\lambda$, stopping potential for photoelectric current is $3 \mathrm{~V}_0$. When the same metallic surface is illuminated with a light of wavelength $2 \lambda$, the stopping potential is $V_0$. The threshold wavelength for the surface is-

UV light of $4.13 \mathrm{~eV}$ is incident on a photosensitive metal surface having work function $3.13 \mathrm{~eV}$. The maximum kinetic energy of ejected photoelectrons will be :

An electron is accelerated through a potential difference of $V$ volt. The speed of electrons will be

Two sources of light emit with a power of $200 {~W}$. The ratio of number of photons of visible light emitted by each source having wavelengths 300 ${nm}$ and $500 {~nm}$ respectively, will be :

Which of the following statement about photon is incorrect?

Ultraviolet light of wavelength $280 \mathrm{~nm}$ is used in an experiment on photoelectric effect with lithium $(\varphi=2.5 \mathrm{~eV})$ cathode. Find

(a) the maximum kinetic energy of the photoelectrons and

(b) the stopping potential.

Which of one is correct

If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is

The wavelength of light in the visible region is about $390 \mathrm{~nm}$ for violet colour. What are the energies of photons in $(\mathrm{eV})$ at the violet end. [Take $h=6.63 \times 10^{-34} \mathrm{~Js}$ and $1 \mathrm{~eV}=1.6 \times 10^{-19} \mathrm{J}]$.

The ratio of the de-Broglie wavelength of an $\alpha$-particle and a proton of same kinetic energy is

In which of the following cases the heavier of the two particle has a same de Broglie wavelength? The two particles :-

When ultraviolet radiation is incident on a surface, no photoelectrons are emitted. If a second beam causes photoelectrons to be ejected, it may consist of

The ratio of the energy of a photon with $\lambda = 150$ nm to that with $\lambda = 300$ nm is

Photoelectric effect can be explained by

The energy of an electron in the ${n^{th}}\,$ Bohr’s orbit is proportional to

If the uncertainty in the position of an electron is ${10^{ - 10}}$m, then the value of uncertainty in its momentum (in kg-$\mathrm{m}{\mathrm{s}^{ - 1}}$) will be

de-Broglie wavelength of a proton and an electron is same. The ratio of kinetic energy of electron to that of proton is

A ultra violet photon and an electron have the same de Broglie wavelengths, then the energy of the photon is -

If ${\mathrm{g}_E}$ and ${\mathrm{g}_M}$ are the acceleration due to gravity on the surfaces of the earth and the moon respectively and it Millikan’s oil drop experiment could be performed on the two surfaces, one will find the ratio $\frac{{\mathrm{electronic\;charge\;on\;the\;moon\;}}}{{\mathrm{electronic\;charge\;on\;the\;earth\;}}}$ to be

A small piece of cesium metal $(\varphi=1.9 \mathrm{~eV})$ is kept at a distance of $20 \mathrm{~cm}$ from a large metal plate having a charge density of $1.0 \times 10^{-9} \mathrm{C} \mathrm{~m}^{-2}$ on the surface facing the cesium piece. A monochromatic light of wavelength $400 \mathrm{~nm}$ is incident on the cesium piece. Find the minimum and the maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present.