Going from K-shell to N-shell in case of H-at | vedclass

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(A) For a hydrogen-like atom,the energy of an electron in the $n^{th}$ shell is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. As the shell numbe

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$K.E.$ decreases

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(A) For a hydrogen-like atom,the energy of an electron in the $n^{th}$ shell is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. As the shell number $n$ increases from $K$ $(n=1)$ to $N$ $(n=4)$,the value of $n$ increases. $1$. $K.E. = |E_n| = 13.6 	imes \frac{Z^2}{n^2}$. As $n$ increases,$K.E.$ decreases. $2$. Total energy $E_n = -13.6 	imes \frac{Z^2}{n^2}$. As $n$ increases,the magnitude of the negative value decreases,meaning the total energy increases. $3$. Potential energy $P.E. = 2 	imes E_n = -27.2 	imes \frac{Z^2}{n^2}$. As $n$ increases,the magnitude of the negative value decreases,meaning the potential energy increases. Therefore,the kinetic energy decreases as the electron moves to a higher shell.

$(P). \lambda_{\max }$ in Lyman series	$(i). \infty \rightarrow 1$
$(Q). v_{\min }$ in Balmer series	$(ii). 2 \rightarrow 1$
$(R). \lambda_{\min }$ in Lyman series	$(iii). 3 \rightarrow 2$
$(S). v_{\max }$ in Balmer series	$(iv). \infty \rightarrow 2$

Going from $K$-shell to $N$-shell in case of $H$-atom :

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If the potential energy of an electron in a hydrogen atom is $-3.02 \ eV$,then the electron is present in which of the following excited states?

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$(P). \lambda_{\max }$ in Lyman series $(i). \infty \rightarrow 1$
$(Q). v_{\min }$ in Balmer series $(ii). 2 \rightarrow 1$
$(R). \lambda_{\min }$ in Lyman series $(iii). 3 \rightarrow 2$
$(S). v_{\max }$ in Balmer series $(iv). \infty \rightarrow 2$

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