Advertisements
Advertisements
Question
State how de-Broglie wavelength (`lambda`) of moving particles varies with their linear momentum (p).
Solution
As, `mc^2 = mc xx c``
`:. E = pc`
`P = E/C`
Since E = hv
`P = (hv)/C`
`P = h/lambda`
or `lambda = h/P`
APPEARS IN
RELATED QUESTIONS
A proton and an α-particle are accelerated through the same potential difference. Which one of the two has less kinetic energy? Justify your answer.
A proton and an electron have same kinetic. Which one has greater de-Broglie wavelength and why?
Show on a graph the variation of the de Broglie wavelength (λ) associated with an electron, with the square root of accelerating potential (V) ?
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?
Let p and E denote the linear momentum and energy of a photon. If the wavelength is decreased,
If the kinetic energy of a free electron double, its de-Broglie wavelength changes by the factor ______.
Two bodies A and B having masses in the ratio of 3 : 1 possess the same kinetic energy. The ratio of linear momentum of B to A is:
For an electron accelerated from rest through a potential V, the de Broglie wavelength associated will be:
The de- Broglie wave length of an electron moving with a speed of 6.6 × 105 m/s is approximately
A proton is accelerated through one volt the increase in its kinetic energy is approximately
