Dual nature of radiation and matter practice · Medium

When radiation of wavelength $\lambda$ is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is

In Millikan’s oil drop experiment, a charged drop of mass $1.8 \times {10^{ - 14}}$ kg is stationary between the plates. The distance between the plates 0.9 cm and potential difference between the plates is 2000 V. The number of electrons on the oil drop is

Statement 1: The photoelectric effect is proof of the quantised nature of light.

Statement 2: Each photon in light beam has same amount of energy

If a proton and electron have the same de Broglie wavelength, then

Cathode rays are

Threshold wavelength for photoelectric effect on sodium is $5000\;\mathop A\limits^ \circ .$ Its work function is

The de Broglie wavelength of an electron having kinetic energy $E$ is $\lambda$. If the kinetic energy of electron becomes $\frac{E}{4}$, then its de-Broglie wavelength will be:

If the work function of a metal is $6.63 \mathrm{~eV}$, then find its threshold frequency for photoelectric effect.

$\alpha -$ particle contain –

The number of photons per second on an average emitted by the source of monochromatic light of wave length 1200 nm, when it delivers the power $3.3 \times {10^{ - 3}}\,\,{\text{watt}}$ will be $\left[ {h = 6.6 \times {{10}^{ - 34}}\,\,J\,\,s} \right]$

Find the de-Broglie wavelength associated with an electron moving with a velocity $0.5c$ and rest mass $=9.1 \times 10^{-31} \mathrm{~kg}$.

The mean free path of the electrons in a discharge tube is 20 cm. The length of the tube is 15 cm only. Then length of Crooke’s dark space is

If wavelength of radiation is $4000 \mathop {\text{A}}\limits^{\text{o}}$, then the energy of the photon is.

In photo electric effect

A. The photocurrent is proportional to the intensity
of the incident radiation.

B. Maximum kinetic energy with which
photoelectrons are emitted depends on the
intensity of incident light.

C. Max. K.E with which photoelectrons are emitted
depends on the frequency of incident light.

D. The emission of photoelectrons require a
minimum threshold intensity of incident
radiation.

E. Max. K.E of the photoelectrons is independent
of the frequency of the incident light.

Choose the correct answer from the options given
below:

Assertion: In photo-particle collision the total energy and total momentum are conserved.

Reason: the number of photos are conserved in a collision.

Assertion: The kinetic energy of photoelectrons emitted from metal surface does not depend on the intensity of incident photon.

Reason: The ejection of electrons from metallic surface is not possible with frequency of incident photons below the threshold frequency.

Assertion (A): Work function of aluminium is 4.2 eV. Emission of electron will not be possible if two photons each of energy 2.5 eV strike an electron of a aluminium.
Reason (R): For photoelectron emission the energy of each photon should be greater than the work function of metal.

Two photon having

In a discharge tube ionization of enclosed gas is produced due to collisions between

The work function for the surface of aluminium is $4.2 \mathrm{~eV}$. What will be the wavelength of that incident light for which stopping potential will be zero? Given $h=6.6 \times 10^{-34} \mathrm{~Js}, c=3 \times 10^8 \mathrm{~ms}^{-1}, 2000 ~\mathop {\text{A}}\limits^ \circ$.

For photoelectric emission from a certain metal, cut off frequency is $v$. If radiation of frequency $2 v$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be ( $m$ is mass of the electron)

Light of wavelength $\lambda$ falls on a metal having work function $\frac{{hc}}{{{\lambda _0}}}.$ Photoelectric effect will take place only if

If de-Broglie wavelength of a moving electron is decreased from $1 \mathop {\text{A}}\limits^{\text{o}}$ to $0.5 \mathop {\text{A}}\limits^{\text{o}}$, then calculate change in its K.E. in eV.

A proton is bound in a nucleus with an energy of 1 MeV. The frequency of radiation required to eject the proton from the nucleus is (in Hz)

The specific charge for positive rays is much less than that for cathode rays. This is because

In a photoelectric experiment using a monochromatic source, on two metals having work functions in ratio 1:2, the de Broglie wavelengths of the fastest photoelectrons in each case are in ratio $1: 2$ if the smaller stopping potential is $0.4 \mathrm{~V}$, the photon energy in $\mathrm{eV}$ is (Nearest integer)

When photon of energy $4.0\,\,eV$ strikes the surface of a metal $A,$ the ejected photoelectrons have maximum kinetic energy ${T_A}\,eV$ and de-Broglie wavelength ${\lambda _A}.$ The maximum kinetic energy of photoelectrons liberated from another metal $B$ by photon of energy $4.50\,\,eV$ is ${T_B} = \left( {{T_A} - 1.5} \right)\,eV.$ If the de-Broglie wavelength of these photoelectrons ${\lambda _B} = 2{\lambda _A},$ then the work function of metal $B$ is

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5 \times {10^{ - 12}}m$, the minimum electron energy required is close to

When a point source of light is at a distance of one metre from a photo cell, the cut off voltage is found to be $V$. If the same source is placed at $2 {~m}$ distance from photo cell, the cut off voltage will be

Assertion: A photon and an electrons, both of energy $1 {MeV}$ has same wavelength.

Reason: $E=2 m_0 c^2=10^6 \mathrm{eV}$