Advertisements
Advertisements
प्रश्न
A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has the greater value of de-Broglie wavelength associated with it, and Give reasons to justify your answer.
उत्तर
de-Broglie wavelength of a particle depends upon its mass and charge for same accelerating potential, such that
`lambdaprop1/sqrt((`
Mass and charge of a proton are mp and e respectively, and, mass and charge of a deutron are 2mp and e respectively. where, e is the charge of an electron
`lambda_p/lambda_D=sqrt((m_Dq_D)/(m_pq_p))=sqrt(((2m_p)(e))/((m_p)(e)))=sqrt2`
Thus, de-broglie wavelength associated with proton is`sqrt2` times of the de-broglie wavelength of deutron and hence it is more.
APPEARS IN
संबंधित प्रश्न
Plot a graph showing variation of de Broglie wavelength λ versus `1/sqrtV` , where V is accelerating potential for two particles A and B, carrying the same charge but different masses m1, m2 (m1 > m2). Which one of the two represents a particle of smaller mass and why?
A proton and an α -particle are accelerated through the same potential difference. Which one of the two has greater de-Broglie wavelength Justify your answer.
State de Broglie hypothesis
Using de Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second postulate of quantization of energy levels in a hydrogen atom.
An electron is accelerated through a potential difference of 64 volts. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond?
Consider the de-Broglie wavelength of an electron and a proton. Which wavelength is smaller if the two particles have (a) the same speed (b) the same momentum (c) the same energy?
Light of wavelength 2000 Å falls on a metal surface of work function 4.2 eV.
What will be the change in the energy of the emitted electrons if the intensity of light with same wavelength is doubled?
When the displacement of a particle executing simple harmonic motion is half of its amplitude, the ratio of its kinetic energy to potential energy is:
A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by 1 m/s so as to have the same K.E. as that of the boy. The original speed of the man will be:
The work function of a metal is 4.50 eV. Find the frequency of light to be used to eject electrons from the metal surface with a maximum kinetic energy of 6.06 × 10−19 J.
