1
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 9th January Morning Slot

For emission line of atomic hydrogen from ni = 8 to nf = n, the plot of wave number $\left( {\overline v } \right)$ against $\left( {{1 \over {{n^2}}}} \right)$ will be (The Rydberg constant, RH is in wave number unit)
A
Linear with intercept $-$RH
B
Non linear
C
Linear with slope RH
D
Linear with slope $-$RH

## Explanation

We know,

$\overline v = {1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_f^2}} - {1 \over {n_i^2}}} \right]$

here ni $=$ 8 and nf $=$ n

Z $=$ 1 for hydrogen

$\therefore$   $\overline v = {1 \over \lambda } = {R_H}\left[ {{1 \over {{n^2}}} - {1 \over {{8^2}}}} \right]$

So, graph is $\to$

So, graph will be linear with slope RH.
2
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 9th January Evening Slot

Which of the following combination of statements is true regarding the interpretation of the atomic orbitals ?

(a) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum.

(b) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number.

(c) According to wave mechanics, the ground state angular momentum is equal to ${h \over {2\pi }}$.

(d) The plot of $\psi$ vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value.
A
(a), (b)
B
(a), (d)
C
(b), (c)
D
(a), (c)
3
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 10th January Evening Slot

The ground state energy of hydrogen atom is – 13.6 eV. The energy of second excited state of He+ ion in eV is :
A
$-$ 6.04
B
$-$ 54.4
C
$-$ 27.2
D
$-$ 3.4

## Explanation

(E)nth = (EGND)H . ${{{Z^2}} \over {{n^2}}}$

E3rd (He+) = ($-$13.6eV) . ${{{2^2}} \over {{3^2}}}$ = $-$ 6.04 eV
4
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 11th January Morning Slot

Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H atom is suitable for this purpose?

[RH = 1 $\times$ 105 cm–1, h = 6.6 $\times$ 10–34 Js, c = 3 $\times$ 108 ms–1]
A
Balmer, $\infty$ $\to$ 2
B
Paschen, 5 $\to$ 3
C
Paschen, $\infty$ $\to$ 3
D
Lyman, $\infty$ $\to$ 1

## Explanation

Given, RH = 1 $\times$ 105 cm–1

$\Rightarrow$ ${1 \over {{R_H}}}$ = 10-5 cm

$\Rightarrow$ ${1 \over {{R_H}}}$ = 10-7 cm $\times$ 100

$\Rightarrow$ ${1 \over {{R_H}}}$ = 100 nm

We know,

${1 \over \lambda } = \nu = {R_H} \times {Z^2}\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)$

$\Rightarrow$ $\lambda$ = ${1 \over {{R_H} \times {{\left( 1 \right)}^2}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$

[For H atom Z = 1]

$\Rightarrow$ $\lambda = {1 \over {{R_H}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$

$\Rightarrow$ $\lambda$ = ${{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$

Given, $\lambda$ = 900 nm

$\therefore$ ${{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$ = 900

$\Rightarrow$ ${\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}$

By checking each options you can see

when nL = 3 and nH = $\infty$ then

${\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}$

$\therefore$ Option C is correct.

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