Question

# The doubly ionized lithium atom is hydrogen-like with the atomic number $$z=3$$. Find the wavelength of the radiation required to excite the electron in $$Li^{2+}$$ from the first to the third Bohr orbit is:[Given:$$\displaystyle \frac{1}{R_{H}}=912 \overset{o}A]$$

A
114oA
B
156oA
C
180oA
D
201oA

Solution

## The correct option is C $$114 \overset{o}A$$Energy $$= 13.6 \ Z^2\left (\cfrac 1{n^2_1}-\cfrac1{n^2_2} \right )$$Given, $$\cfrac 1{R_H}=912 \ \overset{o}A$$$$\cfrac {hc}{\lambda}=13.6 \times Z^2 \left (\cfrac 1{n^2_1}-\cfrac1{n^2_2} \right )$$$$\cfrac 1{\lambda}=R_H \times Z^2 \left (\cfrac 1{n^2_1}-\cfrac1{n^2_2} \right )$$$$n=1$$ to $$n=3$$$$\cfrac 1{\lambda}=R_H \times 9 \left (1- \cfrac 19 \right )$$$$\cfrac 1{\lambda}=8R_H$$$$\Rightarrow \lambda=\cfrac 18 R_H$$$$\lambda = 114 \ \overset{o}A$$Hence, the correct option is $$A$$Chemistry

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