JEE Mains 04-Sep-2020 Shift 2 Solved Paper

As we know ΔE=

hc

λ

So λ=

hc

ΔE

‌‌‌ forλ minimum i.e. shortest;ΔE = maximum
for Lyman series n=1& for ΔEmax
Transition must be form n=∞ to n=1
So

1

λ

=RHZ2(

1

n12

−

1

n22

)

1

λ

=RHZ2(1−0)

1

λ

=R×(1)2⇒λ1=

1

R

For longest wavelength ΔE= minimum for Balmer series n=3 to n=2 will have ΔE minimum
for He+Z=2
So

1

λ2

=RH×Z2(

1

n12

−

1

n22

)

1

λ2

=RH×4(

1

4

−

1

9

)

1

λ2

=RH×

5

9

λ2=λ1×

9

5