Class 12th

12.5 Ground state energy of hydrogen atom, E = -13.6 eV

Kinetic energy is equal to the negative of the total energy (ground state energy) = 13.6 eV

Potential energy is equal to the negative two times of kinetic energy = -13.6 $*2=-27.2eV$

12.4 Separation of two energy level of atom, E = 2.3 eV = 2.3 $*1.6*{10}^{-19}$ J = 3.68 $*{10}^{-19}$

Let $?$ be the frequency of radiation emitted when the atom transits from upper level to lower level.

We have the relation for energy as $E=h?$  , where

h = Planck's constant = 6.626 $*{10}^{-34}$ Js

Then $?=\frac{E}{h}$ = $\frac{3.68*{10}^{-19}}{6.626*{10}^{-34}}$ Hz = 5.55 $*{10}^{14}$ Hz

Hence the frequency is 5.55 $*{10}^{14}$ Hz

12.3 Rydberg's formula is given as:

$\frac{hc}{?}$ = 21.76 $*{10}^{-19}\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

Where, h = Planck's constant = 6.6 $*{10}^{-34}$ Js

c = speed of light = 3 $*{10}^8$ m/s

$?$ = Wavelength

$n_1$ and $n_2$ are integers.

The shortest wavelength present in the Paschen series of the spectral lines is given for the values $n_1=3$ and $n_2=?$

Therefore, $\frac{hc}{?}$ = 21.76 $*{10}^{-19}\left[\frac{1}{3^2}-\frac{1}{{?}^2}\right]$

$\frac{hc}{?}=$ 2.42 $*{10}^{-19}$

$?=\frac{6.6*{10}^{-34}*3*{10}^8}{2.42*{10}^{-19}}$ = 8.189 $*{10}^{-7}$ m= 818.9 nm

12.2 In the alpha-particle scattering experiment, if a thin sheet of hydrogen is used in place of a gold film, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 $*{10}^{-27}$ kg) is less than the mass of incident  $?-$ particles (6.64 $*{10}^{-27}$ kg). Thus, the mass of the scattering particles is more than the target nucleus (hydrogen). As a result, the $?-$ particles would not bounce back if solid hydrogen is used in the $?-$ particle scattering experiment.

12.1 The size of the atom in Thomson's model is no different from the atomic size in Rutherford's model.

In the ground state of Thomson's model, electrons are in stable equilibrium. While in Rutherford's model, electrons always experience a net force.

A classical atom based on Rutherford's model, is doomed to collapse.

An atom has a nearly continuous mass distribution in a Thomson's model, but has a highly non-uniform mass distribution in Rutherford's model.

The positively charged part of the atom possesses most of the mass in both the models.

Diffraction leads to the formation of patterns of varying intensity. When around obstacles, waves bend and spread through the narrow opening, it is called diffraction. The interference results in a new wave pattern and involves the superposition of two or more coherent waves. Both these phenomena produce patterns of light and dark regions; the interference results from the combination of multiple waves and the diffraction arises from a single wave interacting with an aperture or obstacle. When the size of the aperture or obstacle is comparable to the wavelength of the wave, diffraction patterns are typically observed.

A coherent light source in Young's Double Slit Experiment illuminates two closely spaced slits, and produces two overlapping light waves. The interference of these waves constructively or destructively based on their phase difference lead to a pattern of bright and dark fringes on a screen, When the path difference is an integral multiple of the wavelength, it is constructive interference (bright fringes) and when the path difference is an odd multiple of half the wavelength, it is destructive interference (dark fringes). Through observable interference patterns, Young's Double Slit Experiment, shows the wave nature of light.

As per the NCERT Textbooks, Thomson proposed a Atomic Structure of Atom that tells" An atom consists of a positively charged sphere in which the electrons are embedded like the seeds are embedded in watermelon. This model is often compared to a pudding or watermelon with electrons distributed like raisins or plums, also known as “plum pudding model.”

9.5 Actual depth of the bulb in water, $d_1$ = 80 cm = 0.8 m

Refractive index of water, $\mu$ = 1.33

In the given figure, i = angle of incidence, r = angle of refraction = 90 $°$

The light source, bulb (B) is placed at the bottom of the tank.

Since the bulb is a point source, the emergent light can be considered as a circle of radius, with radius R = $\frac{AC}{2}$ = OA = OB

Using Snell's law, we can write the refractive index of water as:

$\mu =\frac{\sin ?r}{\sin ?i}$ = $\frac{\sin ?90°}{\sin ?i}$ = 1.33

i = 48.75 $°$

In Δ OBC, $\tan ?i$ = $\frac{OC}{OB}$

$\tan ?48.75°$ = $\frac{R}{80}$

R = 91.23 cm

Hence, the area of the water surface = $\pi R^2$ = 2.61 $*{10}^4$ ${cm}^2$ = 2.61 $m^2$