The ratio of de Broglie wavelength of a deutron with kinetic energy E to that of an alpha particle with kinetic energy 2E, is n : 1. The value of n is
Q. The ratio of de Broglie wavelength of a deutron with kinetic energy E to that of an alpha particle with kinetic energy 2E, is n : 1. The value of n is ____ .

(Assume mass of proton = mass of neutron) :

Correct Answer: 2

Explanation

The de Broglie wavelength of a particle is given by:

λ = h / √(2mK)

For the deuteron, its mass is equal to the mass of one proton and one neutron:

md = 2m

Its kinetic energy is E, so:

λd = h / √(2 × 2m × E)

For the alpha particle, its mass is equal to two protons and two neutrons:

mα = 4m

Its kinetic energy is 2E, so:

λα = h / √(2 × 4m × 2E)

Now taking the ratio:

λd / λα = √(16mE / 4mE)

λd / λα = √4 = 2

Hence, the value of n is:

2

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