QJEEM14Phf888: JEE Main 2014 Physics - Class 12 JEEM14Phf888 Solution: Hydrogen , Deuterium , singly ionised Helium and doubly ionised …

Question

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Hydrogen

\left( {{}_1{H^1}} \right)

, Deuterium

\left( {{}_1{H^2}} \right)

, singly ionised Helium

{\left( {{}_2H{e^4}} \right)^ + }

and doubly ionised lithium

{\left( {{}_3L{i^6}} \right)^{ + + }}

all have one electron around the nucleus. Consider an electron transition from

n=2

to

n=1.

If the wavelengths of emitted radiation are

{\lambda _1},{\lambda _2},{\lambda _3}

and

{\lambda _4}

respectively then approximately which one of the following is correct?

Accepted Answer

Wave number

{1 \over \lambda } = R{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {{n^2}}}} \right]

\Rightarrow \lambda \propto {1 \over {{Z^2}}}

By question

n=1

and

{n_1} = 2

Then,

{\lambda _1} = {\lambda _2} = 4{\lambda _3} = 9{\lambda _4}

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