Physics Test - 4

$\frac{1}{\lambda }R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$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,${\lambda }_1^{\mathrm{\prime }}=6563$

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

We know that the force on the electron is centripetal in nature.

$F=\frac{k}{r}=\frac{m{v}^2}{r}$

$\Rightarrow v=\sqrt{\frac{k}{m}}$

Now, from Bohr's model, its angular momentum is quantized given by

$L=mvr=\frac{nh}{2\mathrm{\pi }}$

$\Rightarrow r=\frac{nh}{2\mathrm{\pi mv}}$as$v$is constant

$\Rightarrow r\propto n$

For$T$

$T=\frac{1}{2}m{v}^2$

But$v$is independent of$n$

$\Rightarrow T$is independent of$n$

From the conservation of energy the electron kinetic energy equals the maximum photon energy (we neglect the work function$\varphi$because it is normally so small compared to$e{V}_0$).

$e{V}_0=h{v}_{max}$

$e{V}_0=\frac{hc}{{\lambda }_{min}}$

$V_0=\frac{hc}{e{\lambda }_{min}}$

$h$(Plank's Constant)$=6.625\times {10}^{-34}Js,c$(Speed of light)$=3\times {10}^{8}m/s$,$e$(Change of electron)$=1.6\times {10}^{-19}C$

$V_0=\frac{12400\times {10}^{-10}}{{10}^{-11}}$

$=124kV$

Hence, the acceleration voltage for electrons in$X$- ray machine should be less than$124kV$.

$IE=13.6eV$

Excited photon energy$12.1eV$

The energy of the atom at an excited state

$=13.6-12.1=1.5eV$

$\therefore E_n=1.5eV$

$E\propto \frac{1}{n^2}$

$\Rightarrow E_n=\frac{JE}{n^2}$

$\Rightarrow 1.5=\frac{13.6}{n^2}$

$\Rightarrow n^2=\frac{13.6}{1.5}=9$

$\therefore n=3$

As the transition takes from $n=3$ to $n=1$, the energy of the photon emitted is:

$hv=E_3-E_1=\left[\left(-\frac{13.6}{3^2}\right)-\left(-\frac{13.6}{1}\right)\right]=12.1eV$

If$V$is the stopping potential, then$V=\frac{hv-\varphi }{e}$

$V=\frac{12.1-5.1}{e}=7V$

\(\frac{1}{\lambda}=R\left[\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right]\)

In Lyman series, longest wavelength will be emitted if \(n_{i}=2\) and \(n_{f}=1\)

\(\therefore \quad \frac{1}{\lambda_{L}}=R\left(1-\frac{1}{4}\right)=\frac{3 R}{4}\) or \(\lambda_{L}=\frac{4}{3 R}\)

In Balmer series. lonaest wavelenath will be emitted if \(\mathrm{n}_{\mathrm{i}}=3\) and \(\mathrm{n}_{\mathrm{f}}=2\)

\(\therefore \quad \frac{1}{\lambda_{B}}=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5 R}{36}\) or \(\lambda_{B}=\frac{36}{5 R}\)

\(\therefore\) \(\frac{\lambda_{L}}{\lambda_{B}}=\frac{4}{3 R} \times \frac{5 R}{36}=\frac{5}{27}\)