Dual Nature of Radiation · Physics · COMEDK

The basic principle used behind the working of electron microscope is:

When metal of work function 1.4 eV is exposed to a radiation, the maximum kinetic energy of the electron emitted is 0.4 eV . The stopping potential required is:

The variation of the stopping potential ( $\mathrm{V}_{\mathrm{o}}$ ) with the frequency of incident radiation( $n$ ) is as given below.

[no - threshold frequency. h - Planck's constant e-electronic charge] then the slope of the graph AB with the frequency axis is:

What is the ratio of de Broglie wavelength of an electron to that of proton if the velocity of proton is $\frac{1}{6}$ the velocity of electron?

A monochromatic beam of photons of intensity to $1.5 \mathrm{Wm}^{-2}$ and energy 11.2 eV is incident on a material of work function 4.8 eV . What is the maximum speed of photoelectrons emitted due to photoelectric effect if, only $0.53 \%$ of the incident photons eject photoelectrons. Given mass of electron $m=9 \times 10^{-31} \mathrm{~kg}$.

A plot of kinetic energy of emitted photoelectrons from a metal versus the frequency of incident radiation gives a straight line, the intercept of which

A. Depends on the nature of the metal used

B. Depends on the intensity of radiation

C. Depends both on the intensity and the nature of metal used

D. Is a constant and is same for all metals which is independent of the intensity of Incident radiation

A photon emitted during the de-excitation of electron from a state $$\mathrm{n}$$ to the second excited state in a hydrogen atom, irradiates a metallic electrode of work function $$0.5 \mathrm{~eV}$$, in a photocell, with a stopping voltage of $$0.47 \mathrm{~V}$$. Obtain the value of quantum number of the state '$$n$$'.

A particle of mass $$2 \mathrm{mg}$$ has the same wavelength as a neutron moving with a velocity of $$3 \times 10^5 \mathrm{~ms}^{-1}$$. The velocity of the particle is (mass of neutron is $$1.67 \times 10^{-27} \mathrm{Kg}$$)

The velocity of an electron so that its momentum is equal to that of a photon of wavelength $$660 \mathrm{~nm}$$ is

$$\mathrm{K}_1$$ and $$\mathrm{K}_2$$ are maximum kinetic energies of photoelectrons emitted when lights of wavelength $$\lambda_1$$ and $$\lambda_2$$ respectively are incident on a metallic surface. If $$\lambda_1=3 \lambda_2$$, then

The threshold frequency for a metal surface is '$$n_0$$'. A photo electric current '$$I$$' is produced when it is exposed to a light of frequency $$\left(\frac{11}{6}\right) \mathrm{n}_{\mathrm{o}}$$ and intensity $$\mathrm{I}_{\mathrm{n}}$$. If both the frequency and intensity are halved, the new photoelectric current '$$\mathrm{I}^1$$' will become:

The mass of a particle $$\mathrm{A}$$ is double that of the particle $$\mathrm{B}$$ and the kinetic energy of $$\mathrm{B}$$ is $$\frac{1}{8}$$th that of A then the ratio of the de- Broglie wavelength of A to that of B is:

The difference in energy levels of an electron at two excited levels is $$13.75 \mathrm{~eV}$$. If it makes a transition from the higher energy level to the lower energy level then what will be the wave length of the emitted radiation? [given $$h=6.6 \times 10^{-34} \mathrm{~m}^2 \mathrm{~kg} \mathrm{~s}^{-1} ; c=3 \times 10^8 \mathrm{~ms}^{-1} ; 1 \mathrm{~eV}=1.6 \times 10^{-19} \mathrm{~J}$$]

When a certain metal surface is illuminated with light of frequency $$\nu$$, the stopping potential for photoelectric current is $$V_0$$. When the same surface is illuminated by light of frequency $$\frac{\nu}{2}$$, the stopping potential is $$\frac{V_0}{4}$$. The threshold frequency for photoelectric emission is

Let $$K_1$$ be the maximum kinetic energy of photoelectrons emitted by light of wavelength $$\lambda_1$$ and $$K_2$$ corresponding to wavelength $$\lambda_2$$. If $$\lambda_1=2 \lambda_2$$, then

In the photoelectric experiment, the frequency of the incident radiation is doubled. What will be its effect on the photoelectric current?

Ultraviolet light of wavelength 99 mm falls on a metal plate of work function 1.0 eV. If the mass of the electron is 9.1 $$\times$$ 10$$^{-31}$$ kg, the wavelength of the fastest photoelectron emitted is

If K$$_1$$ and K$$_2$$ are maximum kinetic energies of photoelectrons emitted when lights of wavelengths $$\lambda_1$$ and $$\lambda_2$$, respectively incident on a metallic surface and $$\lambda_1=3\lambda_2$$, then

The velocity of the proton is one-fourth the velocity of the electron. What is the ratio of the de-Broglie wavelength of an electron to that of a proton?

From the figure describing photoelectric effect, we may infer correctly that

Light of frequency 10¹⁵ Hz falls on a metal surface of work function 2.5 eV. The stopping potential of photoelectrons (in V) is

A proton accelerated through a potential V has de-Broglie wavelength $$\lambda$$. Then, the de-Broglie wavelength of an $$\alpha$$-particle, when accelerated through the same potential V is