Answer :

a. Before plotting the graph we should know the relation between accelerating potential and de-Broglie wavelength.

According to de-Broglie,

where, p=momentum

h=Planck’s constant

λ=wavelength by the matter

Squaring both sides multiplying on both sides.

.......(1)

As we know the kinetic energy gained by the particle is due to accelerating potential.

The energy gain is given by, .............(2)

where V= accelerating potential

Equating 1 and 2

............. (3)

where c=

This type of equation gives hyperbolic graph,

b. According to question all three particles electron, proton and alpha particle have the same kinetic energy. Hence kinetic energy is constant.

Also, from equation (3)

we can establish that,

Since qV is the gain in kinetic energy so we can write it as K.E

Therefore, now since the electron has the lowest mass, so λ for an electron is maximum followed by proton then alpha particle.

So, has the shortest wavelength.

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