BITSAT 2019 Slot 2 Solved Paper

According to Rydberg’s equation
v=

1

λ

=RZ2[

1

nL2

−

1

nH2

]
where,
v= wave number
λ= wavelength
R= Rydberg constant (=109677cm−1)
Z= atomic number
nL= lower energy state
nH= higher energy state.
i.e.

1

λ

∝Z2
Thus,

λ(He+)

λ(H)

=

Z2(H)

Z2(He+)

To find λ(He)=?
Given, λ(H)=912nm
Z(H)=1
Z(He)=2
∴ λ(He+)=

1×91.2

4

=22.8nm