1

### 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)
2

### 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
3

### 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.
4

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

The de Broglie wavelength ($\lambda$) associated with a photoelectron varies with the frequency (v) of the incident radiation as, [v0 is threshold frequency] :
A
$\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{3 \over 2}}}}}$
B
$\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 4}}}}}$
C
$\lambda \,\infty \,{1 \over {\left( {v - {v_0}} \right)}}$
D
$\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 2}}}}}$

## Explanation

By photoelectric effect

KE = h$\gamma$ - h$\gamma$o ....(1)

de broglie wavelength,

$\lambda$ = ${h \over {mv}}$ = ${h \over {\sqrt {2m \times K.E} }}$ ...(2)

Using equation (1) and (2), we get

$\lambda$ = ${h \over {\sqrt {2m \times \left( {h\nu - h{\nu _0}} \right)} }}$

$\therefore$ $\lambda \,\propto \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 2}}}}}$