1

AIPMT 2012 Prelims

Maximum number of electrons in a subshell with $l$ = 3 and n = 4 is
A
14
B
16
C
10
D
12

Explanation

l = 3 and n = 4 represent 4f, So total number of electrons = 2(2$l$+1) = 14
2

AIPMT 2011 Mains

According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon ?
A
n = 6 to n = 1
B
n = 5 to n = 4
C
n = 6 to n = 5
D
n = 5 to n = 3

Explanation

We know that

$\Delta E \propto \left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$, where n2 > n1

$\therefore$ n = 6 and n = 5 will give least energetic photon.
3

AIPMT 2011 Prelims

If n = 6, the correct sequence for filling of electrons will be
A
$ns \to \left( {n - 2} \right)f \to \left( {n - 1} \right)d \to np$
B
$ns \to \left( {n - 1} \right)d \to \left( {n - 2} \right)f \to np$
C
$ns \to \left( {n - 2} \right)f \to np \to \left( {n - 1} \right)d$
D
$ns \to np\left( {n - 1} \right)d \to \left( {n - 2} \right)f$
4

AIPMT 2011 Prelims

The energies E1 and E2 of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths i.e., $\lambda$1 and $\lambda$2 will be
A
$\lambda$1 = $\lambda$2
B
$\lambda$1 = 2$\lambda$2
C
$\lambda$1 = 4$\lambda$2
D
$\lambda$1 = ${1 \over 2}$ $\lambda$2

Explanation

E1 = ${{hc} \over {{\lambda _1}}}$ and E2 = ${{hc} \over {{\lambda _2}}}$

${{{E_1}} \over {{E_2}}}$ = ${{hc} \over {{\lambda _1}}} \times {{{\lambda _2}} \over {hc}}$ = ${{{\lambda _2}} \over {{\lambda _1}}}$

$\Rightarrow$ $25 \over 50$ = ${{{\lambda _2}} \over {{\lambda _1}}}$

$\Rightarrow$ $1 \over 2$ = ${{{\lambda _2}} \over {{\lambda _1}}}$ $\Rightarrow$ $\lambda _1$ = 2$\lambda _2$