UPSEE 2018 Solved Paper

As, the expression for the de-Broglie wavelength associated with a charged particle having charge Q and mass m is
λ=

h

√2mqV

where, V= potential through which the particle is accelerated.
Now, given
m1=m2=m
and

q1

q2

=

1

2

⇒q2=2q1
Then,

λ1

λ2

=

h √2mqv

h √4mqV

=

√2

1

Hence, λ1=√2λ2
⇒λ1>λ2