Physics Test - 17

Given:

\(n_{1}=4\) and \(n_{2}=5\)

The wavelength of the radiations emitted from the hydrogen atom is given by:

\(\frac{1}{\lambda_{1}}=R\left[\frac{1}{(4)^{2}}-\frac{1}{(5)^{2}}\right]\)

\(= R\left[\frac{1}{16}-\frac{1}{25}\right]=\frac{9 R}{400}\).....(1)

When electrons jump from \(4^{\text {th }}\) to \(1^{\text {st }}\) orbit, then the wavelength of the radiations emitted from the hydrogen atom is given by:

\(\frac{1}{\lambda_{2}}=R\left[\frac{1}{(1)^{2}}-\frac{1}{(4)^{2}}\right]\)

\(= R\left[\frac{1}{1}-\frac{1}{16}\right]=\frac{15 R}{16}\).....(2)

Divide equation (1) and (2), we get,

\(\frac{\lambda_{1}}{\lambda_{2}}=\frac{400 \times 15 R}{9 R \times 16}\)

\(=\frac{125}{3}\)

\(=125 : 3\)

Magnetic moment \(\mathrm{M}=\) Current \(\times\) Area \(=\frac{\text { Charge } \times \text { Area }}{\text { Time }}\)

\(=\frac{q \times \pi r^{2}}{T}=\frac{1}{2} q \omega r^{2} \quad\left(\because \omega=\frac{2 \pi}{T}\right)\)

Angular momentum \(L=m \omega r^{2}\)

\(\therefore \frac{M}{L}=\frac{\frac{1}{2} q \omega r^{2}}{m \omega r^{2}}=\frac{q}{2 m}\)

Thus, the ratio of its magnitude of its magnetic moment to that of its angular momentum depends on \(\mathrm{q}\) and \(\mathrm{m}\)

Given,

\({u}=50 {~km} / {hr}=50 \times \frac{5}{18}=\frac{250}{18} {~m} / {s}\)

\({v}=0\)

\({s}=6 {~m} \quad\) (at least)

By third equation by motion,

\({v}^{2}={u}^{2}+2 {as}\)

\(0=\left(\frac{250}{18} \times \frac{250}{18}\right)+2(-{a})({s})\)

\(0=192.90-12 {a}\)

\({a}=\frac{192.90}{12}=16.07\)

Now applying third equation of motion,

\({v}^{2}={u}^{2}+2 \mathrm{as} \)

\(0=\left(100 \times \frac{5^{2}}{18}\right)+2(-16.07)({s}) \)

\(0=771.60-32.14 {~s} \)

\({s}=\frac{771.60}{32.14}=24.00\)

\(\therefore\) The minimum stopping distance is \(24\) m.

The energy \(E\) of the emitted photon is given by:

\(E=\frac{h c}{\lambda}\)...(i)

Here, \(h\) is Planck's constant, \(c\) is the speed of light and \(\lambda\) is the wavelength of the photon emitted.

The energy \(E_{n}\) of the electron in the nth energy level is given by

\(E_{n}=-\frac{13.6}{n^{2}} {eV}\)...(ii)

The change in angular momentum \(\Delta L\) of an atom is given by

\(\Delta L=\left(n_{1}-n_{2}\right) \frac{h}{2 \pi} \ldots \ldots(iii)\)

Here, \(n_{1}\) is the initial energy level, \(n_{2}\) is the final energy level and \(h\) is the Planck's constant.

We have given that the wavelength of the photon emitted by the hydrogen atom is \(1027 \stackrel{\circ}{{A}} . \lambda=1027 \stackrel{\circ}{{A}}\).

Substitute \(6.626 \times 10^{-34} {~J} \cdot {s}\) for \(h, 3 \times 10^{8} {~m} / {s}\) for \(c\) and \(1027 \stackrel{0}{{~A}}\) for \(\lambda\) in equation (i)

\(E=\frac{\left(6.626 \times 10^{-34} {~J} \cdot {s}\right)\left(3 \times 10^{8} {~m} / {s}\right)}{1027 \stackrel{\circ}{{A}}}\)

\(E=\frac{\left(6.626 \times 10^{-34} {~J} \cdot {s}\right)\left(3 \times 10^{8} {~m} / {s}\right)}{1027 \times 10^{-10} {~m}}\)

\(E=19.35 \times 10^{-19} {~J}\)

Convert this value of energy of the hydrogen atom into electronvolt.

\(\Delta E=\frac{19.35 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{C}} \)

\(\Delta E=12.1 {eV}\)

Hence, the energy of the photon emitted from the hydrogen atom is \(12.1 \mathrm{eV}\).

Let us calculate the energy difference between the first and third energy level of the hydrogen atom using equation (ii).

\(E_{3}-E_{1}=\left(-\frac{13.6}{(3)^{2}} {eV}\right)-\left(-\frac{13.6}{(1)^{2}} {eV}\right)\)

\(E_{3}-E_{1}=(-1.5 {eV})-(-13.6 {eV})\)

\(E_{3}-E_{1}=12.1 {eV}\)

From the above calculations, it is clear that the electron in the hydrogen atom jumps from third energy level to the first energy level.Let us now calculate the change in angular momentum of the hydrogen atom using equation (3).

Substitute 3 for \(n_{1}\) and 1 for \(n_{2}\) in equation (3).

\(\Delta L=(3-1) \frac{h}{2 \pi} \)

\(\Delta L=(2) \frac{h}{2 \pi}\)

\(\Delta L=\frac{h}{\pi}\)

Therefore, the change in angular momentum of the hydrogen atom is \(\frac{h}{\pi}\)

Amplitude of an damped oscillation is given using the equation, \(A=A_{0} \cdot e^{\frac{-b t}{2 m}}\). It is known that energy of the damped oscillation system is directly related to its amplitude. It is said that since the system dissipates energy gradually with respect to time, the amplitude will decrease with time.

Now, work done is defined as a product of force and displacement. Here, the force acting on the mass is due to spring. Hence, \(W=F \times x^{2}\), which means that work done is directly proportional to square of displacement. In graphical terms, displacement in waveforms is measured as amplitude. Thus energy is proportional to square of amplitude.

When energy is halved, \(\frac{E}{2} \propto a_{\text {new}}{ }^{2}\). This implies

When energy is halved, \(\frac{E}{2} \propto a_{n e w}^{2}\). This implies

\(\sqrt{\frac{E}{2}} \propto a_{n e w}\)

\(\frac{a_{0}}{\sqrt{2}} \propto a_{n e w}\)

Substituting the value in the amplitude equation we get,

\(A_{n e w}=A_{0} \cdot e^{\frac{-b t}{2 m}}\)

\(\frac{A_{0}}{\sqrt{2}}=A_{0} \cdot e^{\frac{-b t}{2 m}}\) (since we know that \(A_{n e w}\frac{A_{0}}{ \sqrt{2}}\) which was derived above)

Cancelling the like term we get

\(\frac{1}{\sqrt{2}}=e^{\frac{-b t}{2 m}}\)

We know that damping constant \(b\) is \(10^{-2}\) and mass \(m\) as \(0.1 kgs\). Substituting this we get,

\(\frac{1}{\sqrt{2}}=e^{\frac{-10^{-2} t}{2 \times 0.1}}\)

Removing \(e\) by multiplying lnon both sides we get,

\(\ln \left(\frac{1}{\sqrt{2}}\right)=\frac{-10^{-2} t}{0.2}\)

\(-0.35=\frac{-10^{-2} t}{0.2}\)

\(0.35 \times 20=t\)

\(t=7 s\)

A simple pendulum has a string with a very small mass (sphere ball). It is small but strong enough not to stretch. The mass suspended from the light string that can oscillate when displaced from its rest position.

Pendulums are in common usage such as in clocks, child’s swing and some are for fun. For small displacements, a pendulum is a simple harmonic oscillator.

The pendulum’s time period is proportional to the square root of the length of string and inversely proportional to the square root of acceleration due to gravity.

The time period of a simple pendulum is given by

\(T \propto \sqrt{\frac{l}{g}}\)

\(T \propto \sqrt{\frac{1}{g}} ; T \propto \sqrt{l}\)

Where,

T is the time period

l is the length of the pendulum

g is the acceleration due to gravity

Let us take the relation,

\(T \propto \sqrt{l}\)

\(T \propto l \frac{1}{2}\)

Given that,

The relative difference in the periods, is \(5 \times 10^{-4} s\)

\(\Delta T=5 \times 10^{-4} s\)

The pendulum is attached to different spherical bobs of radii \(r_{1}\) and \(r_{2}\)

The change in length is the difference in radii

\(\Delta l=\left|r_{1}-r_{2}\right|\)

Substitute the above in equation 1

\(\Delta T=\frac{1}{2} \Delta l \rightarrow 1\)

\(5 \times 10^{-4}=\frac{1}{2}\left|r_{1}-r_{2}\right|\)

\(\left|r_{1}-r_{2}\right|=2 \times 5 \times 10^{-4}\)

\(\left|r_{1}-r_{2}\right|=10 \times 10^{-4} m\)

\(\left|r_{1}-r_{2}\right|=0.1 \times 10^{-2} m\)

\(\left|r_{1}-r_{2}\right|=0.1 cm\)

The difference in radii, \(\left|r_{1}-r_{2}\right|\) is \(0.1 cm\)

Given,

Mass of the body, \(m=5.0 \) kg

Magnitude of forces,

\(\left|\overrightarrow{F_{1}}\right|=8\)Newton

\(\left|\overrightarrow{F_{2}}\right|=6\) Newton

The magnitude of the resultant force of these forces,

\(F=\sqrt{\left|\overrightarrow{F_{1}}\right|^{2}+\left|\overrightarrow{F_{2}}\right|^{2}}\)

\(=\sqrt{\left[8^{2}+6^{2}\right]}\)

\(=\sqrt{(64+36)}\)

\(=\sqrt{100}=10\) Newton

Magnitude of acceleration,

\(a=\frac{F}{M}\)

\(=\frac{10}{5.0}\)

\(=2{ m/s^2}\)

Magnetic Flux \(\phi=\vec{B} \cdot \vec{A}\)

\(\Rightarrow \text { Magnetic flux }(\phi)=\text { B.A.Cos } \theta\)

Given Magnetic Field \(B=\frac{1}{\pi}\left(\frac{{Wb}}{{m}^{2}}\right)\)

Area enclosed \((\) Circular \()=\pi r^{2}\)

\(r=0.2 m\)

Area \(=\pi(0.2)^{2}\)

Angle between field and area \(\theta=60^{\circ}\)

Magnetic Flux \(\phi= \text{B.A.Cos} \theta=\frac{1}{\pi}\left(\frac{W b}{m^{2}}\right)\left(\pi(0.2)^{2}\right)\left(\frac{1}{2}\right)=0.02 Wb\)

Given that:

Dipole Moment, \(P=3 \times 10^{-3} \mathrm{~cm}\)

Electric Field Intensity, \(E=10^{4} \mathrm{NC}^{-1}\)

We know that:

In rotating the dipole from the position of stable equilibrium by an angle \(\theta\), the amount of work done is given by,

\(W=P E(1-\cos \theta)\)

For unstable equilibrium, \(\theta=180^{\circ}\)

\(\therefore W=P E\left(1-\cos 180^{\circ}\right) \quad\left[\because \cos 180^{\circ}=-1\right] \)

\(=2 P E \)

\(=2 \times 3 \times 10^{-3} \times 10^{4} \mathrm{~J} \)

\(=60 \mathrm{~J}\)