The energy of an electron in the ground state (n=1) for He^ + ion is -x J , then that for an electron in n=2 state for Be^ 3+ ion in J is

Correct answer: D. -x E_n=-R_H((z^2)/(n^2)) J For He^ + (n=1) , aligned & E_n=-x=-R_H((2^2)/(1^2))=-4 R_H & R_H=(x)/(4) & aligned For Be^ 3+ (n=2) , aligned E_n & =-R_H((z^2)/(n^2)) J & =-(x)/(4) ×((4 × 4)/(2 × 2))=-x J aligned

NEET 2024

Exam: NEET
Year: 2024
Subject: Chemistry
Chapter: Structure of Atom

NEET 2024 · Chemistry · Structure of Atom

The energy of an electron in the ground state (n=1) for He^ + ion...

The energy of an electron in the ground state

(n=1)

for

\mathrm{He}^{+}

ion is

-x J

, then that for an electron in

n=2

state for

\mathrm{Be}^{3+}

ion in J is

Options

A $-\frac{x}{9}$
B $-4 x$
C $-\frac{4}{9} \mathrm{x}$
D $-x$

Correct answer

D. $-x$

Step-by-step solution

E_n=-R_H\left(\frac{z^2}{n^2}\right) J

For

\mathrm{He}^{+}(n=1)

\begin{aligned} & E_n=-x=-R_H\left(\frac{2^2}{1^2}\right)=-4 R_H \\ & \therefore \quad R_H=\frac{x}{4} \\ & \end{aligned}

For

\mathrm{Be}^{3+}(n=2)

\begin{aligned} E_n & =-R_H\left(\frac{z^2}{n^2}\right) J \\ & =-\frac{x}{4} \times\left(\frac{4 \times 4}{2 \times 2}\right)=-x J \end{aligned}

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