JEE MAIN - Physics (2014 (Offline) - No. 6)
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?
$$4{\lambda _1} = 2{\lambda _2} = 2{\lambda _3} = {\lambda _4}$$
$${\lambda _1} = 2{\lambda _2} = 2{\lambda _3} = {\lambda _4}$$
$${\lambda _1} = {\lambda _2} = 4{\lambda _3} = 9{\lambda _4}$$
$${\lambda _1} = 2{\lambda _2} = 3{\lambda _3} = 4{\lambda _4}$$
Explanation
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, $${\lambda _1} = {\lambda _2} = 4{\lambda _3} = 9{\lambda _4}$$
$$ \Rightarrow \lambda \propto {1 \over {{Z^2}}}$$
By question $$n=1$$ and $${n_1} = 2$$
Then, $${\lambda _1} = {\lambda _2} = 4{\lambda _3} = 9{\lambda _4}$$
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