Dual Nature of Radiation · Physics · COMEDK

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 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