Physics Test - 48

Energy in nth orbit of Hydrogen atom is given by:

\(E_{n}=-\frac{13.6 Z^{2}}{n^{2}} e V\)

Where \(Z\) is the atomic number and \(n\) is the principal quantum number.

The velocity of the revolving electron is:

\(v=\frac{e^{2}}{2 n h \epsilon_{0}}\)

The Rydberg's constant is:

\(R=\frac{m e^{4}}{8 \epsilon_{0}^{2} h^{3} c}\)

Thus,

\(v R=\frac{e^{2}}{2 n h \epsilon_{0}} \times \frac{m e^{4}}{8 \epsilon_{0}^{2} h^{3} c}=\frac{m e^{6}}{16 n h^{4} \in_{0}^{3} c}\)

Since, \(\mathrm{n}, \mathrm{h}, \epsilon_{0}, \mathrm{~m}, \mathrm{c}, \mathrm{e}\) are constant.

So, the VR is inversely proportional to the principal quantum number (n).

The energy of an electron is:

\(E=\frac{-m e^{4}}{8 n^{2} h^{2} C_{0}^{2}}\)

Thus,

\(\frac{v}{E}=\frac{\frac{c^{2}}{2 n h \epsilon_{0}}}{\frac{-m_{e^{4}}}{8 n^{2} h^{2} \in_{0}^{2}}}\)

\(\frac{v}{E}=\frac{e^{2}}{2 n h \epsilon_{0}} \times \frac{8 n^{2} h^{2} \epsilon_{0}^{2} v}{-m e^{4} E}=-\frac{4 n h \epsilon_{0}}{m e^{2}}\)

Since, \(\mathrm{n}, \mathrm{h}, \epsilon_{0}, \mathrm{~m}\) and \(\mathrm{e}\) are constant.

So, \(\frac{v}{E}\) is proportional to principal quantum number \(\mathrm{n}\).

Both upper half and lower half will have the same effective area of \(\frac{\pi R^{2}}{2}\) so a change in flux will be same and induced emf will have the same value. But since the resistance is different due to which current must be different but the ring is as a whole is closed circuit so the electric field will be generated to make the current flow in both parts to be same.

Gravitational force between two objects of mass \(m_{1}\) and \(m_{2}\) separated by a distance \(r, F=\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)

\[\Rightarrow \mathrm{G}=\frac{F \mathrm{r}^{2}}{\mathrm{~m}_{1} \mathrm{~m}_{2}}\]

Value of universal gravitational constant \(\mathrm{G}\) is \(6.67 \times 10^{-11} \frac{\mathrm{Nm}^{2}}{\mathrm{~kg}^{2}}\).

Relation between electric field \(\mathrm{E}_0\) and magnetic field \(\mathrm{B}_0\) of an electromagnetic wave is given by

\(\mathrm{c}=\frac{\mathrm{E}_0}{\mathrm{~B}_0}(\text { Here }, \mathrm{c}=\text { Speed of light }) \)

\(\Rightarrow \mathrm{E}_0=\mathrm{B}_0 \times \mathrm{c}=1.2 \times 10^{-7} \times 3 \times 10^8=36\)

As the wave is propagating along \(x\)-direction, magnetic field is along \(z\)-direction

\(\text { and }(\hat{\mathrm{E}} \times \hat{\mathrm{B}}) \| \hat{\mathrm{C}}\)

\(\therefore \mathrm{E}\) should be along y -direction.

So, electric field \(\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \sin \overrightarrow{\mathrm{E}} \cdot(\mathrm{x}, \mathrm{t})\)

\(=\left[-36 \sin \left(0.5 \times 10^3 \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \hat{\mathrm{j}}\right] \frac{\mathrm{V}}{\mathrm{m}}\)

Given,

Thickness of the dielectric slab, \(t=1 cm =10^{-2} m\)

Dielectric constant, \(\varepsilon_{\tau}=K=5\)

Area of the plates of the capacitor, \(A=0.01 m ^{2}=10^{-2} m ^{2}\)

Distance between parallel plates of the capacitor, \(d=2 \) cm =\(2 \times 10^{-2} m\)

We know that:

Capacity with air in between the plates,

\(C_{0}=\frac{\epsilon_{0} A}{d}\)

where, \(\epsilon_{0}=8.854 \times 10^{-12}\)

\(=\frac{8.85 \times 10^{-12} \times 10^{-2}}{2 \times 10^{-2}} \)

\(C_{0}=4.425 \times 10^{-12} Farad\)

Capacity with dielectric slab in between the plates,

\(C=\frac{\epsilon_{0} A}{d-t\left(1-\frac{1}{K}\right)} \)

\(=\frac{8.85 \times 10^{-12} \times 10^{-2}}{\left(2 \times 10^{-2}\right)-10^{-2}\left(1-\frac{1}{5}\right)} \)

\(C=7.375 \times 10^{-12} \text { Farad }\)

Increase in capacity on introduction of dielectric:

\(C-C_{0}=\left(7.375 \times 10^{-12}\right)-\left(4.425 \times 10^{-12}\right) \)

\(=2.95 \times 10^{-12} \text { Farad }\)

An element produces bright and dark lines with the same wavelengths.

For example, hydrogen has three prominent lines with wavelengths of 434 nm, 486 nm, and 656 nm.

The emission spectrum appears dark if the hydrogen is absorbing light, and bright if it is emitting light, but the same three wavelengths are seen in either case.

Each element produces unique bright line spectra because no two elements have the same number of electrons.

So, the wavelength in the bright-line emission spectrum of an element is characteristic of the particular element.

When moving through a magnetic field, the charged particle experiences a force.

When the direction of the velocity of the charged particle is perpendicular to the magnetic field:

• Magnetic force is always perpendicular to velocity and the field by the Right-Hand Rule.
• And the particle starts to follow a curved path.
• The particle continuously follows this curved path until it forms a complete circle.
• This magnetic force works as the centripetal force.

Centripetal force (F_C) = Magnetic force (F_B)

​q v B = $\frac{m{v}^2}{R}$

R = $\frac{mv}{qB}$

When a charge moves through a magnetic field in a perpendicular direction, it will move in a circular motion as explained above with a radius 

​R = $\frac{mv}{qB}$

​For alpha particle according to the question mass is m, velocity is v, the charge is 2e, and the magnetic field is B. 

So,

$R=\frac{mv}{qB}=\frac{mv}{\left(2e\right)B}$

The amplitude of the waves are:

\(\begin{aligned} a_{1} &=10 \mathrm{\mu m} \\ a_{2} &=4 \mathrm{\mu m} \\ a_{3} &=7 \mathrm{\mu m} \end{aligned}\)

The Phose differenve between \(1^{st}\) and \(2^{\text {nd }}\) wave is \(\frac{\pi}{2}\) and that between \(2^{\text {nd }}\) and \(3^{rd}\) wave is \(\frac{\pi}{2}\). Therefore , phose differente between \(1^{st}\) ond \(3^{r d}\) is \(\pi\). Combining \(1^{st}\) and \(3^{rd}\) their resultant amplitude is given by:

\(A_{1}^{2} =a_{1}^{2}+a_{3}^{2}+2 a_{1} a_{3} \cos \phi \)

\(\text { or } A_{1} =\sqrt{10^{2}+7^{2}+2 \times 10 \times 7 \cos \phi} \)

\(=\sqrt{100+49-140} \)

\(=\sqrt{9}\)

\(=3 \mathrm{\mu m}\)

Now Combining this with \(2^{\text {nd }}\) wave we have, the resultant amplitude:

\(A^{2} =A_{1}^{2}+a_{2}^{2}+2 A_{1} a_{2} \cos \frac{\pi}{ 2} \)

\( = \sqrt{3^{2}+4^{2}+2 \times 3 \times 4 \cos 90^{\circ}}\)

\(=\sqrt{9+16} \)

\(=5 ~\mu m\)

The potential energy of hydrogen can be given by:

\(\mathrm{P} . \mathrm{E} .=-\frac{\mathrm{k} Z_{\mathrm{e}}^{2}}{\mathrm{r}}\)

As the atom is excited, its radius increases. But due to the negative sign Its actual potential energy increases.

And Kinetic energy is given by

\(\mathrm K . \mathrm{E} .=\frac{\mathrm{k} Z_{\mathrm{e}}^{2}}{2 \mathrm{r}}\)

As the atom is excited, its radius increases thus resulting in a decrease in Kinetic energy.

When a hydrogen atom is raised from the ground state, it increases the value of r (distance from centre).As the distance of the electron from the nucleus increases.

As a result of this,P.E. increases(decreases in negative) andK.E. decreases.