CBSE Class 12 · 2020 · Modern PhysicsMultiple Choice

WhatsApp Email

Photons of energies 1 eV and 2 eV are successively incident on a metallic surface of work function 0.5 eV. The ratio of kinetic energy of most energetic photoelectrons in the two cases will be:

(A)1 : 2
(B)1 : 1
(C)1 : 3✓ Correct
(D)1 : 4

Answer: (C)

Solution

K_{ma x} = h ν - ϕ_{0}

.
Case 1:

K_{1} = 1 - 0.5 = 0.5 eV

. Case 2:

K_{2} = 2 - 0.5 = 1.5 eV

\frac{K _{1}}{K _{2}} = \frac{0.5}{1.5} = \frac{1}{3} .

(C) 1 : 3

CBSE Class 12 Board · 2020 · Single correct option

Practise more: all CBSE board PYQs on Modern Physics →

More board questions on Modern Physics

The potential energy between two nucleons inside a nucleus is minimum at a distance of about
CBSE 2024View solution →
The variation of the stopping potential $V_{0}$ with the frequency $ν$ of the incident radiation for four metals A, B, C and D is shown in the figure. For the same frequency of incident radiation producing photo-electrons in all metals, the kinetic energy of photo-electrons will …
CBSE 2024View solution →
The energy of an electron in the ground state of a hydrogen atom is $- 13.6$ eV. The kinetic and potential energy of the electron in the first excited state will be
CBSE 2024View solution →
Assertion (A): An alpha particle is moving towards a gold nucleus. The impact parameter is maximum for the scattering angle of $18 0^{\circ}$ .
Reason (R): The impact parameter in an alpha particle scattering experiment does not depend upon the atomic number of the target nucleus.
CBSE 2024View solution →
Calculate the energy released / absorbed (in MeV) in the nuclear reaction $_{1}^{1} H +_{1}^{3} H \to_{1}^{2} H +_{1}^{2} H$ . Given $m (_{1}^{1} H) = 1.007825 u$ , $m (_{1}^{2} H) = 2.014102 u$ , …
CBSE 2024View solution →
A proton of energy 1.6 MeV approaches a gold nucleus $(Z = 79)$ . Find the distance of its closest approach.
CBSE 2024View solution →