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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
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B
156oA
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C
180oA
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D
201oA
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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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