1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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$$\lambda _2$$

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