# De Broglie’s relation MCQs with Answers

## De Broglie’s relation MCQs

According to de Broglie’s hypothesis, moving has electrons with momentum P associated to wavelength can be given as,

∴ λ = h/p

Where, h is Planck’s constant, P is momentum of electron.

If an electron of mass m is accelerated by a potential difference V, the work has done on electron increases its KE.  The energy of electron is given by.

E (Works done) = eV Q.1) Whose results are used by de Broglie’s to prove his hypothesis?

a) Einstein and Plank

b) Einstein and Maxwell

c) Plank and Maxwell

d) None of these

Q.2) According to de Broglie’s hypothesis, the wavelength of electron is………..

a) directly proportional to velocity of particle

b) directly proportional to square velocity of particle

c) inversely proportional to velocity of particle

d) inversely proportional to square of velocity of particle

Q.3) According to de Broglie’s relation if velocity of particle is infinite, wavelength will be…

a) infinite

b) small

c) large

d) zero

Q.4) Momentum of particle of radiating energy with wavelength of 1.5 A0is,

a) 1.4 ×10-14 kg-m/s

b) 2.4 ×10-14 kg-m/s

c) 3.4 × 10-14 kg-m/s

d) 4.4 ×10-14 kg-m/s Q.5) Speed of electron emitted with wavelength of 5000 Ao will be

a) 1.457 ×105 m/s

b)1.457 ×106  m/s

c) 1.457 ×107 m/s

d) 1.457 ×108 m/s Q.6) Filament emits electrons when potential difference of 50 V is applied across it, wavelength of electron is…….

a) 1.63 A⁰

b) 1.73 A⁰

c)1.83 A⁰

d) None of these Q.7) Wavelength of 1.65 A0 is emitted by filament in Davisson- Germer experiment, the magnitude of applied potential difference will be

a) 35 V

b) 45 V

c) 55 V

d) 65 V Q.8) Momentum of electrons in terms of potential difference can be given as,

a) P = 1/√2meV

b) P = √meV

c) P = 1/√meV

d) P = √2meV

Q.1) a) Einstein and Plank

Q.2) c) inversely proportional to velocity of particle

Q.3) d) zero

Q.4) d) 4.4 ×10-14  kg-m/s

Q.5) b)1.457 ×106  m/s

Q.6) b) 1.73 A⁰

Q.7) c) 55 V

Q.8) d) P = √2meV

Updated: July 22, 2021 — 6:36 pm