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$\lambda_{e} = \frac{12.27}{\sqrt{49}} \implies \frac{12.27}{7} = 1.75\text{ \AA}$
$\lambda = \frac{h}{\sqrt{2mqV}}$
$\lambda = \left(\frac{h}{\sqrt{2mq}}\right)\frac{1}{\sqrt{V}}$
Slope of $(\lambda - 1/\sqrt{V})$ graph $= h/\sqrt{2mq}$
or slope $\propto 1/\sqrt{m}$
$\phi_{0} = h\nu_{0} = \frac{hc}{\lambda_{0}} \implies \lambda_{0} \propto \frac{1}{\phi_{0}}$
$KE_{max} = (4 - 2)\text{ eV}$
hence cut off potential $eV_{0} = 2\text{ eV} \implies V_{0} = 2\text{ volt}$
Detailed solution not required.
$\lambda = \frac{h}{\sqrt{3mk_{B}T}}$
$= \frac{6.6 \times 10^{-34}}{\sqrt{3 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times 400}}$
$= 1.254 \times 10^{-10}\text{ m} = 1.254\text{ \AA}$
Also:
$\lambda = \frac{h}{P}$
$\frac{d\lambda}{\lambda} = -\frac{dP}{P} \implies \frac{0.4}{100} = \frac{P}{P} \implies P^{\prime} = 250P$
$P_{1} = P_{2}$
$\frac{h}{\lambda_{1}} = \frac{h}{\lambda_{2}}$
$\lambda_{1} = \lambda_{2}$
$n \propto \frac{1}{d^{2}}$
$\frac{n_{1}}{n_{2}} = \left(\frac{d_{2}}{d_{1}}\right)^{2}$
$\frac{n_{1}}{n_{2}} = \left(\frac{0.5}{1.0}\right)^{2}$
$\implies n_{2} = 4n_{1}$
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Detailed solution not required.
Series limit of Lyman series is $\infty \rightarrow 1$ (frequency $\nu_{1}$)
First line of Lyman series is $2 \rightarrow 1$ (frequency $\nu_{2}$)
Series limit for Balmer series is $\infty \rightarrow 2$ (frequency $\nu_{3}$)
$h\nu_{3} + h\nu_{2} = h\nu_{1}$
$\nu_{1} - \nu_{2} = \nu_{3}$
$\frac{1}{\lambda} = R\left[\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right]$
$\implies \frac{1}{970.6 \times 10^{-10}} = 1.097 \times 10^{7}\left[\frac{1}{1^{2}} - \frac{1}{n_{2}^{2}}\right]$
$\implies n_{2} = 4$
Number of emission lines $N = \frac{n(n-1)}{2} = \frac{4 \times 3}{2} = 6$
$\frac{1}{\lambda} = R\left(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right)$
$\frac{1}{\lambda_{0}} = R\left(\frac{1}{2^{2}} - \frac{1}{3^{2}}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$
$\frac{1}{\lambda} = R\left(\frac{1}{2^{2}} - \frac{1}{4^{2}}\right) = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}$
$\frac{\lambda}{\lambda_{0}} = \frac{5}{36} \times \frac{16}{3} = \frac{20}{27} \implies \lambda = \frac{20}{27}\lambda_{0}$
$(r_{n})_{He^{+}} = \frac{(r_{n})_{H}}{Z} = \frac{0.53\text{ \AA}}{2} = 0.265\text{ \AA}$
Also, $(E_{n})_{He^{+}} = Z^{2}(E_{n})_{H}$. ($Z = 2$ for $\text{He}^{+}$)
Detailed solution not required.
K.E. of electron $= |E| = 3.4\text{ eV}$
$= 3.4 \times 1.6 \times 10^{-19}\text{ J} = 5.44 \times 10^{-19}\text{ J}$
de-Broglie wavelength, $\lambda = \frac{h}{\sqrt{2mE}}$
$= \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 5.44 \times 10^{-19}}}$
$= 6.658 \times 10^{-10}\text{ m} = 6.658\text{ \AA}.$
The energy of an electron in the nth orbit of hydrogen atom is $E_{n} \propto \frac{m}{n^{2}}$
For the hypothetical particle whose mass is double than the electron, energy is $E_{n} \propto \frac{2m}{n^{2}}$
The wavelength emitted by the transition from $n_{i}$ to $n_{f}$ state is $\frac{1}{\lambda^{\prime}} = 2R\left[\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}\right]$
Here $n_{f} = 2$ [First excited state]
Longest wavelength will be emitted if transition is from $n_{i} = 3$ to $n_{f} = 2$.
$\frac{1}{\lambda^{\prime}} = 2R\left[\frac{1}{2^{2}} - \frac{1}{3^{2}}\right] = 2R\left[\frac{1}{4} - \frac{1}{9}\right] = \frac{5R}{18}$
$\lambda^{\prime} = \frac{18}{5R}$
Here, $r_{0} = 4 \times 10^{-14}\text{ m}$, $E = 10\text{ MeV} = 10 \times 1.6 \times 10^{-13}\text{ J}$, $e = 1.6 \times 10^{-19}\text{ C}$
Using relation $r_{0} = \frac{1}{4\pi\epsilon_{0}} \cdot \frac{2Ze^{2}}{E}$,
$Z = \frac{4\pi\epsilon_{0} \times r_{0}E}{2e^{2}} = \frac{4 \times 10^{-14} \times 1.6 \times 10^{-12}}{9 \times 10^{9} \times 2 \times (1.6 \times 10^{-19})^{2}} = 139$
The nuclear reaction is: ${}_{92}^{236}X \rightarrow {}_{a}^{141}Y + {}_{36}^{b}Z + 3 {}_{0}^{1}n + Q$
As in a nuclear reaction both Z and A no. remain conserved.
$92 = a + 36 \implies a = 56$
$236 = 141 + b + 3 \implies b = 92$
The Q-value of the reaction is given by:
$Q = (m_{x} - m_{y} - m_{z} - 3m_{n})\text{ a.m.u.}$
$= (235.891 - 140.673 - 91.791 - 3 \times 1.009)\text{ a.m.u.}$
$= 0.4\text{ a.m.u.} = 0.4 \times 931.5\text{ MeV}.$
The attractive nuclear force is the same for any pair of nucleons.
Thus, $F_{1} = F_{3}$ when there are no electrostatic forces. But $F_{2} =$ attractive nuclear force - repulsive electrostatic force.
Hence, $F_{1} = F_{3} > F_{2}.$
Detailed solution not required.
In 1st excited state $n = 2$.
Velocity of electron is $\frac{1}{2}(\frac{c}{137})$.
Wavelength of series limit of lyman:
$\frac{1}{\lambda} = R\left(\frac{1}{1^{2}} - \frac{1}{\infty^{2}}\right)$
$\lambda = \frac{1}{R}$
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$\frac{1}{\lambda} \propto Z^{2} \implies \frac{\lambda_{2}}{\lambda_{1}} = \frac{Z_{1}^{2}}{Z_{2}^{2}}$
$\lambda_{2} = \left(\frac{1}{3}\right)^{2} \times \lambda_{1} = \frac{\lambda}{9}$
Detailed solution not required.
$V_{net} = 5.5 - 0.4 = 5.1\text{ V}$
$R_{net} = 5.1 \times 10^{3}~\Omega$
$i = \frac{5.1}{5.1 \times 10^{3}}$
$i = 1\text{ mA}.$
Reverse biased potential for the Zener breakdown.
$V_{r} = Ed = 10^{6} \times 2.5 \times 10^{-6} - 1$
$= 2.5\text{ V} - 1\text{ V} = 1.5\text{ V}$
Detailed solution not required.
Avg. of voltage or mean voltage $= \frac{V_{0}}{\pi} = \frac{10}{\pi}$
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