AP EAPCET 19-Aug-2021 Shift 1 Paper

Let us consider, particle A having velocity =vA
and particle B having velocity =vB
According to de-Broglie wavelength, λ=

h

mv

here, h= Planck constant
m= mass of electron
v= velocity of particle
Now, λA=

h

mAvA

and λB=

h

mBvA

Wavelength of A is found to be double than of B.
i.e. λA=2×λB ...(i)
Put value of λA and λB in Eq. (i)
2×

h

mBVB

=

h

mAVA

‌‌‌‌‌‌‌‌ (For vA=vB)

2

mB

=

1

mA

‌‌‌‌‌‌ [ mA=

mB

2

]
Here, mass of A is half that of B.