Class 12th

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation-A set of atoms in an excited state decays in general to any of the states with lower energy.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation -Total energy in ground state collide elastically Is 213.6eV= 27.2eV

When a electron excite from ground state to first excited state then

Energy is =-13.6/2²= 3.4eV+13.6 (ground state energy) eV = 17.0 eV

So net energy is 27.2-17 =10.2eV

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explantion- in the molecules, nuclear force between the nuclei of the two atoms is not important because nuclear forces are short-ranged.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a)

Explanation- In the simple Bohr model, only the magnitude of angular momentum is kept equal to some integral multiple of h/2 π, where, h is Planck's constant and thus, the Bohr model gives incorrect values of angular momentum

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation-The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because of the electrons not being subject to a central force.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation- when one decides to work in a frame of reference where the electron is at rest, the given expression is not true as it the non-inertial frame of reference.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (c)

Explanation- The atomic number of lithium is 3, therefore, the radius of Li+ ion in its ground state, on the basis of Bohr's model, will be about 1/3 times to that of Bohr radius.

Therefore, the radius of lithium ion is near 53/3=18≈ pm.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Hγ in Balmer series corresponds to transition n= 5 to n= 2. So, the electron in ground state i.e., from n= 1must first be placed in state n= 5.

Energy required for the transition from n= 2 to n= 5 is given by E₁-E₅=13.6-0.54=13.06eV

Since, angular momentum is conserved,

angular momentum corresponding to Hg photon= change in angular momentum of electron

so L₅-L₂=5h-2h= 3h = 3 $*1.06*{10}^{-34}$

=3.18 $*{10}^{-34}$ kgm²/s

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- The frequency of any line in a series in the spectrum of hydrogen like atoms corresponding to the transition of electrons from (n+ p) level to nth level can be expressed as a difference of two terms;

From the formula we can say that v=cRZ² { $\frac{1}{({n+p)}^2}-\frac{1}{n^2}$ }

Where p =1,2,3, … R os rydberg constant

When p<

V=cRZ² { $\frac{1}{n^2}({1+\frac{p}{n})}^{-2}-\frac{1}{n^2}$ }

V=cRZ² { $\frac{1}{n^2}-\frac{2p}{n^3}-\frac{1}{n^2}\}$

By binomial theorem

V=cRZ² { $\frac{2p}{n^3}=\frac{2cR{Z}^2}{n^3}p$

Thus, the first few frequencies of light that is emitted when electrons fall to the nth level

from levels higher than n, are approximate harmonic (i.e., in the ratio 1 : 2 : 3 …) when

n >> 1 .

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- The electron in Hydrogen atom in ground state revolves on a circular path whose radius is equal to the Bohr radius (a_n) . Let the velocity of electron is v

∴ Number of revolutions per unit time= $\frac{2\pi a_0}{V}$

The electric current is given by i=q/t, if q charge flows in time t. Here,   q=e

The electric current is given by i= $\frac{2\pi a_0}{V}e$