AP EAPCET 25-Aug-2021 Shift 1 Paper

According to de-Broglie's wavelength, λ=

h

mv

Now, de-Broglic's wavelength of particle A is
λA=

h

mAvA

Similarly, de-Broglie's wavelength of particle B is
λA=

h

mBvB

∴ Ratio of de-Broglic wavelength of A and B is 2:1
i.e.,

λA

λB

=

h

mAVA

×

mBvB

h

2

1

=

h

mA×0.05

×

mB×0.02

h

2

1

=

mB×0.02

mA×0.05

2

0.02

=

mB

mA×0.05

1

0.01

=

mA

mB×0.05

⇒

mA

mB

=

1

5

Hence, the ratio of their masses mA:mB is 1:5.