Physics

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.

6.22 Mass lifted, m = 10 kg

Height to which the mass lifted, h = 0.5 m

No of repetitions, n = 1000

(a) Work done against gravitational force,

W = nmgh = 1000 $*10*9.81*0.5=$ 49050 J

(b) Mechanical energy supplied by 1 kg fat, with 20% efficiency rate = 0.2 $*$ 3.8 $*{10}^7$ = 0.76 $*{10}^7$ J/kg

Fat used by dieter = 49050 / (0.76 $*{10}^7)$ kg = $6.45*{10}^{-3}$ kg

6.21 Given, the area of the windmill sweep = A, Wind velocity = v

The volume of air passing through the blade = Av

Let the density of air be $?$ , the mass of air passing through the blade = $?Av$

(a) The mass of air passing through the blade in time t = $?Avt$

(b) The kinetic energy of air = $\left(\frac{1}{2}\right)m{v}^2$ = $\left(\frac{1}{2}\right)?Avt{v}^2$ = $\left(?At{v}^3\right)$ /2 …. (1)

(c) Area, A = 30 $m^2$ , v = 36 km/h = 10 m/s, density of air be $?$ = 1.2 kg/ $m^3$

Total wind energy, from eqn. (1) = 18 kW

Electrical energy = 25 % of wind energy = 0.25 $*18=4.5kW$

6.20 Mass of the body = 0.5 kg

Velocity, v = a x 3/2

a = 5 m–1/2 s–1

At x =0, the initial velocity, u = 0

At x = 2, the final velocity, v = 5 $*2^{3/2}$ = 14.142 m/s

Work done by the system = increase in K.E. of the body = (1/2)m ( $v^2$ - $u^2$ )

= (1/2) $*0.5*$ 14.142 $*14.142$ = 50 J

6.19 Given, the mass of the trolley,  $m_t$ = 300 kg, mass of the sand bag,  $m_s$ = 25 kg, uniform velocity of the trolley, v = 27 kmph = 0.75 m/s

Since there is no external force acting on the system, the speed of the trolley will remain unchanged even after entire sand is empty. 27 kmph is the answer.

6.18 The length of the pendulum, l = 1.5 m

The potential energy of the bob at horizontal position = mgl

Since it dissipates 5% of its kinetic energy to come to the horizontal position, from the law of conservation of energy we get,

$\left(\frac{1}{2}\right)m{v}^2$ = (0.95) $*mgl$

$v^2$ = 2 $*0.95*9.81*1.5$

v = 5.287 m/s

6.17 In an elastic collision, when the ball A will hit the ball B, A comes to rest immediately and the ball B acquires the velocity of ball A. The momentum thus gets transferred from a moving body to a stationary body.