A proton and an alpha particle are accelerated through the same potential. Which one of the two has (i) greater value of de-Broglie wavelength associated with it, and (ii) less kinetic energy? Justify your answer.

Solution

(i) The de-Broglie wavelength associated with same potential V is,

λ = \frac{h}{p} or λ = \frac{h}{mv}

∴

λ \propto \frac{1}{\sqrt{m}}

Wavelength and mass has inverse dependence a seen from the above relation.

As proton's mass is less than the mass of alpha particle thus,

λ_{proton} > λ_{alpha} .

Proton has greater value of de-broglie wavelength assosciated with it.

(ii) Now,
Energy of a photon is, E = h

ν

i.e.,

K . E . = \frac{hc}{λ}

\Rightarrow

K . E . \propto \frac{1}{λ}

Since,

λ_{proton} > λ_{alpha .}

[from (i) ]

Thus, kinetic energy of proton will be lesser than that of alpha particle.

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