Physics Test - 2

Proton is accelerated through a potential difference of  $100 V$100V

So, $\begin{array}{l}\frac{1}{2}m{v}^2=e\times 100\\ mv=\sqrt{2m_{p}e\times 100}\\ {\lambda }_o=\frac{h}{\sqrt{2me\times 100}}\end{array}$

De Broglie wavelength of $\alpha$− particle

$\begin{array}{l}{\lambda }_{\alpha }=\frac{h}{\sqrt{2m_{\alpha }{v}_{\alpha }\times 100}}\\ {\lambda }_{\alpha }=\frac{h}{\sqrt{2\times 4m\times 2e\times 100}}\end{array}$

So, ${\lambda }_{\alpha }=\frac{{\lambda }_0}{2\sqrt{2}}$

$\frac{1}{\lambda }R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ for Balmer series

Put $\lambda$ in the above equation we get $R{Z}^2=1.097\times {10}^7$

For the second member

$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

$n_1=2$ and $n_2=3, 4, 5...$

$\frac{1}{\lambda }=1.097\times {10}^7\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$

$\lambda =4861.5 \AA$

From Bohr's atomic model, the atomic radius for $n^{th}$ state is given by the formula,

The wavelength of the spectral lines of the hydrogen spectrum is given by the formula:

$\frac{1}{\lambda }=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

Where $R=$ Rydberg constant

For the first member of the Balmer series, $n_f=2$ and $n_i=3$

$\therefore \frac{1}{{\lambda }_1}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$

$\therefore \frac{1}{{\lambda }_1}=\frac{5R}{36}$...(1)

For the first member of the Lyman series, $n_f=1$ and $n_i=2$

$\therefore \frac{1}{{\lambda }_1^{\text{'}}}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$

$\therefore \frac{1}{{\lambda }_1^{\text{'}}}=\frac{3R}{4}$...(2)

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

$\begin{array}{l}\frac{{\lambda }_1}{{\lambda }_1^{\mathrm{\prime }}}=\frac{5R}{36}\times \frac{4}{3R}\\ \therefore \frac{{\lambda }_1}{{\lambda }_1^{\mathrm{\prime }}}=\frac{5}{27}\\ \therefore {\lambda }_1=\frac{5}{27}{\lambda }_1^{\mathrm{\prime }}\\ \end{array}$

Given that, ${\lambda }_1^{\mathrm{\prime }}=6563$

$\begin{array}{l}\therefore {\lambda }_1=\frac{5}{27}\times 6563\\ \therefore {\lambda }_1=1215.4 \AA \end{array}$

As given that proton has a specific charge to be $9.6\times {10}^7 C k{g}^{-1}$

Now as $\alpha$ particle is the Helium nucleus, so it must have $+2$ unit charge.

Specific charge of proton $=\frac{specific charge of proton}{mass of proton in particle}$

$\alpha$ particle is doubly ionized $He$ atom. It carries $+2e$ charge [Twice the charge of proton and mass of $\alpha$ particle is ($2$ Neutrons $+2$ Protons) is almost $4$ times the mass of proton].

So, if the Specific charge of proton $=\frac{+e}{m_p}$

Then, a Specific charge of $\alpha$ particle $=\frac{+2e}{4m_p}=$ Specific charge of proton $\times \frac{1}{2}$

So, the Specific charge of $\alpha$ particle $=\frac{9.6\times {10}^7}{2}=4.8\times {10}^7 C k{g}^{-1}$

A transformer is used to bring the voltage up or down in an AC electrical circuit.

Transformers are used in various fields like power generation grid, distribution sector, transmission and electric energy consumption. There are various types of transformers which are classified based on the following factors:-

• Working voltage range.
• The medium used in the core.
• Winding arrangement.
• Installation location.

For the first-line of the Lyman series,

$\begin{array}{l}{n}_1=1 \text{and }{n}_2=3\\ \therefore \frac{1}{{\lambda }_1}=R(\frac{1}{1^2}-\frac{1}{2^2})\\ =R(1-\frac{1}{4})\\ \end{array}$
$=\frac{3R}{4}$...(1)
For first line of Paschen series

$\begin{array}{l}{n}_1=3 \text{ and }{n}_2=4\\ \therefore \frac{1}{{\lambda }_2}=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)\\ =R\left(\frac{1}{9}-\frac{1}{16}\right)\\ \end{array}$

$=\frac{7R}{144}$...(2)

From equation (1) and (2)

$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{7R}{144}\times \frac{4}{3R}$

$=\frac{7}{108}$