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The longest wavelength of light that can be used for the ionisation of \text{Li}^{2+} ion in its ground state is \text{x} \times 10^{-8} \mathrm{~m}. The value of x is____________ . (Nearest Integer)

(Given : Energy of the electron in the first shell of the hydrogen atom is -2.2\times10^{-18}\text{J}; \mathrm{h=6.63\times10^{34}Js \; and \; c=3\times10^{8}ms^{-1}})

Option: 1

1


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} \mathrm{I E=E_{0} \times \frac{Z^{2}}{n^{2}} }&=2.2 \times 10^{-18} \times 9 \\ &=19.8 \times 10^{-18} \mathrm{~J} \end{aligned}

\begin{aligned} &\therefore \frac{h c}{\lambda}=19.8 \times 10^{-18}\\ &\Rightarrow \lambda=\frac{h c}{19.8 \times 10^{-18}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{19.8 \times 10^{-18}}\\ \\&\; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;=1 \times 10^{-8} \end{aligned}

Hence, the answer is 1

Posted by

Suraj Bhandari

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