The ratio of de-Broglie wavelength of an α-particle and a proton accelerated from rest by the same potential is 1m√ , the value of m is:
Question
Solution 1
The de-Broglie wavelength (λ) is given by the formula:
λ = h / p
where h is Planck's constant and p is the momentum of the particle.
The momentum of a particle can be expressed in terms of its kinetic energy (K.E.) as:
p = √(2mK.E.)
where m is the mass of the particle.
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