Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Dual Nature of Radiation question

2020 · 2 Sep · Shift 2 · Q49
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Dual Nature of Radiation
  5. /2020 · 2 Sep · Shift 2 · Q49

Dual Nature of Radiation question

2020 · 2 Sep · Shift 2 · Q49

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is 1.878 ×\times× 10–4. The mass of the particle is close to
  1. A
    1.2 ×\times× 10–28 kg
  2. B
    9.1 ×\times× 10–31 kg
  3. C
    4.8 ×\times× 10–27 kg
  4. D
    9.7 ×\times× 10–28 kg
View written solutionFree

Correct answer: D

  1. Use de-Broglie wavelength formula

For a particle,

λ=hmv\lambda = \frac{h}{mv}λ=mvh​

So for the given particle and electron,

λpλe=h/(mpvp)h/(meve)=mevempvp\frac{\lambda_p}{\lambda_e} = \frac{h/(m_p v_p)}{h/(m_e v_e)} = \frac{m_e v_e}{m_p v_p}λe​λp​​=h/(me​ve​)h/(mp​vp​)​=mp​vp​me​ve​​
  1. Use the speed relation

The particle is moving 5 times as fast as the electron:

vp=5vev_p = 5v_evp​=5ve​

Hence,

λpλe=mevemp(5ve)=me5mp\frac{\lambda_p}{\lambda_e} = \frac{m_e v_e}{m_p (5v_e)} = \frac{m_e}{5m_p}λe​λp​​=mp​(5ve​)me​ve​​=5mp​me​​

Given,

λpλe=1.878×10−4\frac{\lambda_p}{\lambda_e} = 1.878 \times 10^{-4}λe​λp​​=1.878×10−4

Therefore,

1.878×10−4=me5mp1.878 \times 10^{-4} = \frac{m_e}{5m_p}1.878×10−4=5mp​me​​

So,

mp=me5(1.878×10−4)m_p = \frac{m_e}{5(1.878 \times 10^{-4})}mp​=5(1.878×10−4)me​​
  1. Substitute electron mass

Using

me=9.1×10−31 kgm_e = 9.1 \times 10^{-31}\,\text{kg}me​=9.1×10−31kg mp=9.1×10−315×1.878×10−4m_p = \frac{9.1 \times 10^{-31}}{5 \times 1.878 \times 10^{-4}}mp​=5×1.878×10−49.1×10−31​

First compute denominator:

5×1.878×10−4=9.39×10−45 \times 1.878 \times 10^{-4} = 9.39 \times 10^{-4}5×1.878×10−4=9.39×10−4

Thus,

mp=9.1×10−319.39×10−4m_p = \frac{9.1 \times 10^{-31}}{9.39 \times 10^{-4}}mp​=9.39×10−49.1×10−31​ mp=(9.19.39)×10−27m_p = \left(\frac{9.1}{9.39}\right) \times 10^{-27}mp​=(9.399.1​)×10−27 mp≈0.969×10−27m_p \approx 0.969 \times 10^{-27}mp​≈0.969×10−27 mp≈9.69×10−28 kgm_p \approx 9.69 \times 10^{-28}\,\text{kg}mp​≈9.69×10−28kg
  1. Match with options
mp≈9.7×10−28 kgm_p \approx 9.7 \times 10^{-28}\,\text{kg}mp​≈9.7×10−28kg

So the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

This matches the derived answer.

PreviousNext

More from Dual Nature of Radiation

  • When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to :2020 · MCQ
  • Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of photons of X-rays to the number density of…2020 · MCQ
  • Given figure shows few data points in a phot electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is: (Plancks constant h = 6.62 × 10–34 J.s) Includes diagram2020 · MCQ
  • Particle A of mass mA = 2m​ moving along the x-axis with velocity v0 collides elastically with another particle B at rest having mass mB =3m​. If both particles move along the x-axis after the collision, the change Δλ…2020 · MCQ
  • In a photoelectric effect experiment, the graph of stopping potential V versus reciprocal of wavelength obtained is shown in the figure. As the intensity of incident radiation is increased : Includes diagram2020 · MCQ
  • The surface of a metal is illuminated alternately with photons of energies E1 = 4 eV and E2 = 2.5 eV respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is ​…2020 · Numerical
  • An electron, a doubly ionized helium ion (He++) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths λ e, λ He++ and λ p is :2020 · MCQ
  • Assuming the nitrogen molecule is moving with r.m.s. velocity at 400 K, the de-Broglie wavelength of nitrogen molecule is close to : (Given : nitrogen molecule weight : 4.64 × 10–26 kg, Boltzman constant: 1.38 × 10–23 J/K,…2020 · MCQ