NEET 2024 — Chemistry solution

NEET2024Chemistry

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:

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

Solution

Bohr energy: $E_{n} \propto - \frac{Z ^{2}}{n ^{2}}$ .

For He $^{+}$ ( $Z = 2, n = 1$ ): $E = - k \cdot 4 = - x \Rightarrow k = x /4$ .

For Be $^{3 +}$ ( $Z = 4, n = 2$ ): $E = - k \cdot \frac{16}{4} = - 4 k = - x$ .

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