KEAM 2018 Physics and Chemistry Paper

Given,‌

EA

EB

=‌

1

2

,‌

UA

UB

=‌

1

2

So EA=x,EB=2x
and UA=y,UB=2y
∵‌‌EA=UA+KA
and EB=UB+KB
here KA and KB are kinetic energy of particles A and B
So KA=EA−UA=(x−y).....(i)
KB=EB−UB=2(x−y)....(ii)
∵ de-Broglie wavelength,
λ=‌

h

√2mk

So ‌‌λA=‌

h

√2mKA

,λB=‌

h

√2mKB

∴‌‌‌

λA

λB

=√‌

KB

KA

......(iii)
From Eq. (i), (ii) and (iii),
‌

λA

λB

=√‌

2(x−y)

(x−y)

=‌

√2

1