IIT JEE Advanced 2006 Paper 1

We have

1

λ

= RZ2(1−n12−

1

n22

)
For nth line of Lyman series for hydrogen-like atom (Z = 11), we have

1

λ

= R(1−

1

h2

)×112
The de Broglie wavelength is
λ =

h

mv

=

hr

mvr

=

hr

nh∕2π

Therefore,
λ =

2πr

n

=

2π

n

×

(0.53n2)

z

Å
where Z = 11. Hence,

1

λ

=

11

2π×0.53×10−10n

= 1.1 × 107(1−

1

n2

)×112
Therefore,
n -

1

n

=

1

2π×0.53×10−4×112

= 25
n2 – 1 = 25 m
n2 – 25n – 1 = 0
n =

25+√625+4

2

n = 25