1
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
2
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.
3
MCQ (Single Correct Answer)

### 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

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

$\therefore$ $\lambda \,\propto \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 2}}}}}$
4
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 12th January Morning Slot

What is the work function of the metal if the light of wavelength 4000$\mathop A\limits^ \circ$ generates photoelectrons of velocity 6 $\times$ 105 ms–1 from it ?
(Mass of electron = 9 $\times$ 10–31 kg;
Velocity of light = 3 $\times$ 108 ms$-$1
Plank's constant = 6.626 $\times$ 10–34 Js;
Charge of electron = 1.6 $\times$10–19 JeV–1)
A
4.0 eV
B
0.9 eV
C
2.1 eV
D
3.1 eV

## Explanation

E = $\phi$ + K.E

$\Rightarrow$ h$\nu$ = $\phi$ + ${1 \over 2}m{v^2}$

$\Rightarrow$ $\phi$ = $h\nu - {1 \over 2}m{v^2}$

= ${{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {4000 \times {{10}^{ - 10}}}} - {1 \over 2} \times 9 \times {10^{ - 31}} \times {\left( {6 \times {{10}^5}} \right)^2}$

= 3.35 $\times$ 10-19 J

$\Rightarrow$ $\phi$ = ${{3.35 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}$ eV

= 2.0934 eV $\simeq$ 2.1 eV

### EXAM MAP

#### Joint Entrance Examination

JEE Advanced JEE Main

#### Graduate Aptitude Test in Engineering

GATE CSE GATE EE GATE ECE GATE ME GATE CE GATE PI GATE IN

NEET