1

JEE Advanced 2021 Paper 2 Online

Numerical
Consider a helium (He) atom that absorbs a photon of wavelength 330 nm. The change in the velocity (in cm s$$-$$1) of He atom after the photon absorption is __________.

(Assume : Momentum is conserved when photon is absorbed.

Use : Planck constant = 6.6 $$\times$$ 10$$-$$34 J s, Avogadro number = 6 $$\times$$ 1023 mol$$-$$1, Molar mass of He = 4 g mol$$-$$1)
Your Input ________

Answer

Correct Answer is 30

Explanation

Wavelength of photon absorbed, $$\lambda$$ = 330 nm = 330 $$\times$$ 10$$-$$9 m

Planck's constant, h = 6.6 $$\times$$ 10$$-$$34 J s

Molar mass of He, M = 4 g mol$$-$$1 = 4 $$\times$$ 10$$-$$3 kg mol$$-$$1

Avogadro number, NA = 6 $$\times$$ 1023 mol$$-$$1

Mass of one atom of He, $$m = {M \over {{N_A}}}$$

$$ = {{4 \times {{10}^{ - 3}}} \over {6 \times {{10}^{23}}}} = {2 \over 3} \times {10^{ - 26}}$$ kg

Velocity, = V cm/s.

Using de-Broglie equation,

$$\lambda = {h \over {mv}}$$

$$ \Rightarrow v = {h \over {m\lambda }} = {{6.6 \times {{10}^{ - 34}}} \over {2/3 \times {{10}^{ - 26}} \times 330 \times {{10}^{ - 9}}}}$$

$$ = {{6.6 \times 3 \times {{10}^{ - 34}} \times {{10}^{35}}} \over {2 \times 330}} = 0.03 \times 10 = 0.3$$ m/s

= 30 cm/s
2

JEE Advanced 2020 Paper 2 Offline

Numerical
The figure below is the plot of potential energy versus internuclear distance (d) of H2 molecule in the electronic ground state. What is the value of the net potential energy E0 (as indicated in the figure) in kJ mol-1, for d = d0 at which the electron-electron repulsion and the nucleus-nucleus repulsion energies are absent? As reference, the potential energy of H atom is taken as zero when its electron and the nucleus are infinitely far apart.
Use Avogadro constant as 6.023 $$ \times $$ 1023 mol-1.

Your Input ________

Answer

Correct Answer is $$-$$5242.41

Explanation

Given that, electrons and nucleus are at infinite distance, so potential energy of H-atom is taken as zero.

Therefore, according to Bohr's model, potential energy of a H-atom with electron in its ground state = $$-$$27.2 eV

At d = d0, nucleus-nucleus and electron-electron repulsion is absent.

Hence, potential energy will be calculated for 2 H atoms = $$-$$2 $$ \times $$ 27.2 eV = $$-$$54.4 eV

Potential energy of 1 mol H atoms in kJ

= $${{54.4 \times 6.02 \times {{10}^{23}} \times 1.6 \times {{10}^{ - 19}}} \over {1000}}$$

= $$ - 5242.4192$$ kJ/mol
3

JEE Advanced 2015 Paper 1 Offline

Numerical
Not considering the electronic spin, the degeneracy of the second excited state( n = 3) of H atom is 9, while the degeneracy of the second excited state of H is
Your Input ________

Answer

Correct Answer is 3
4

JEE Advanced 2014 Paper 1 Offline

Numerical
In an atom, the total number of electrons having quantum numbers n = 4, |ml| = 1 and ms = –1/2 is
Your Input ________

Answer

Correct Answer is 6

Explanation

As n = 4, so l = 0, 1, 2, 3 which implies the orbitals are 4s, 4p, 4d and 4f.

Now, |ml | = 1 implies ml = +1 and −1. Therefore, l can be 3, 2, 1, as for l = 0, ml = 0.

For l = 1, ml = −1, 0, +1

For l = 2, ml = −2, −1, 0, +1, +2

For l = 3, ml = −3, −2, −1, 0, +1, +2, +3

Therefore, there are six possible orbitals with |ml | = 1 . Also, given that ms = −1/2 so six electrons are possible with ms = −1/2.

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

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

Medical

NEET

CBSE

Class 12