Numerical Problems - Chapter 8 Atomic and Nuclear Physics 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
What is the distance of closest approach when a $5 \mathrm{MeV}$ proton approaches a gold nucleus.
Solution:
$
\begin{aligned}
& \mathrm{q}_1=\mathrm{ze} \\
& \mathrm{q}_2=\mathrm{e}
\end{aligned}
$
At the distance $r_0$ of closest approach,
K.E of a Proton $=$ P.E. of proton and the gold nucleus
$
\begin{aligned}
\mathrm{K}=\frac{1}{2} m v^2 & =\frac{1}{4 \pi \varepsilon_o} \frac{Z e . e}{r_o} \\
r_0 & =\frac{1}{4 \pi \varepsilon_o} \frac{Z e^2}{\mathrm{~K}}
\end{aligned}
$
But $\mathrm{K}=5 \mathrm{MeV}=5 \times 1.6 \times 10^{-13} \mathrm{~J}$
For gold, $Z=79$
$
\begin{aligned}
r_{\mathrm{o}} & =\frac{9 \times 10^9 \times 79 \times\left(1.6 \times 10^{-19}\right)^2}{5 \times 1.6 \times 10^{-13}} \\
& =2.28 \times 10^{-14} \mathrm{~m} \\
r_{\mathrm{o}} & =2.3 \times 10^{-14} \mathrm{~m}
\end{aligned}
$

Question 2.
Calculate the impact parameter of a $5 \mathrm{MeV}$ particle scattered by $90^{\circ}$ when it approaches.
Solution:
$
\begin{aligned}
& \mathrm{KE}=5 \mathrm{MeV}=5 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J} \\
& \theta=90^{\circ}
\end{aligned}
$
For gold, $\mathrm{Z}=79$
$
\begin{aligned}
b & =\frac{\mathrm{KZ} e^2 \cot \left(\frac{\theta}{2}\right)}{\mathrm{KE}} \\
b & =\frac{9 \times 10^9 \times 79 \times\left(1.6 \times 10^{-19}\right)^2 \times \cot 45^{\circ}}{5 \times 1.6 \times 10^{-13}} \\
& =2.27 \times 10^{-14} \mathrm{~m} ; \quad b=2.3 \times 10^{-14} \mathrm{~m}
\end{aligned}
$
Impact parameter,
Question 3.
What is the angular momentum of an electron in the third orbit of an atom?
Solution:
Here $\mathrm{n}=3 ; \mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}$
Angular momentum,
$
\mathrm{L}=\frac{n h}{2 \pi}=\frac{3 \times 6.6 \times 10^{-34}}{2 \times 3.14}=3.15 \times 10^{-34} \mathrm{JS}
$

Question 4.
Write down the expression for the radii of orbits of hydrogen atom. Calculate the radius of the smallest orbit.
Solution:
The radius of the $\mathrm{n}^{\text {th }}$ orbit of a hydrogen atom is given by
$
\mathrm{r}=\frac{n^2 h^2}{4 \pi^2 m K e^2}
$
Radius of innermost orbit, called Bohr's radius, is obtained by putting $\mathrm{n}=1$. It is denoted by $r_0$
$
\begin{aligned}
& r_{\mathrm{o}}=\frac{h^2}{4 \pi^2 m K e^2}=\frac{\left(6.6 \times 10^{-34}\right)^2}{4 \times(3.14)^2 \times 9.1 \times 10^{-31} \times 9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2} \\
& \mathrm{r}_0=0.53 \times 10^{-10} \mathrm{~m}=0.53 \mathrm{~A}^{\circ} .
\end{aligned}
$
Question 5.
Calculate the frequency of the photon, which can excite the electron to $-3.4 \mathrm{eV}$ from -13.6 eV.
Solution:
Energy of photon, hv $=\mathrm{E}_2-\mathrm{E}_1$
$
\begin{aligned}
\text { Frequency, } v & =\frac{\mathrm{E}_2-\mathrm{E}_1}{h} \\
& =\frac{[-3.4-(-13.6)] \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} \\
& =\frac{10.2 \times 1.6 \times 10^{15}}{6.6} \\
v=2.47 \times 10^{15} \mathrm{~Hz} &
\end{aligned}
$

Question 6.
The ground state energy of hydrogen atom is $-13.6 \mathrm{eV}$. If an electron makes a transition from an energy level $-0.85 \mathrm{eV}$ to $-1.51 \mathrm{eV}$, Calculate the wavelenth of the spectral line emitted. To which series of hydrogen spectrum does this wavelenth belong?
Solution:
$
\begin{aligned}
& \text { Here } \Delta \mathrm{E}=\mathrm{E}_2-\mathrm{E}_1=-0.85-(-1.51) . \\
& =0.66 \mathrm{eV} \\
& \Delta \mathrm{E}=0.66 \times 1.6 \times 10^{-19} \mathrm{~J} \\
& \lambda=\frac{h c}{\Delta E}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{0.66 \times 1.6 \times 10^{-19}} \\
& =18.84 \times 10^{-7} \\
& \lambda=18840 \AA
\end{aligned}
$
This wavelength belongs to the Pachen series of the hydrogen spectrum.
Question 7.
Express 16 rag mass into equivalant energy in $\mathrm{eV}$.
Solution:
Here $\mathrm{m}=16 \mathrm{mg}=16 \times 10^{-16} \mathrm{~kg}, \mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1}$
Equivalent energy, $\mathrm{E}=\mathrm{mc}^2$
$
\begin{aligned}
& =16 \times 10^{-16} \times\left(3 \times 10^8\right)^2 \mathrm{~J} \\
& =\frac{16 \times 10^{-6} \times\left(3 \times 10^8\right)^2}{1.6 \times 10^{-19}} \mathrm{eV} \\
& \mathrm{E}=9 \times 10^{30} \mathrm{eV} .
\end{aligned}
$

Question 8.
The nuclear mass of ${ }_{26}^{56} \mathrm{Fe}$ is $55.85 \mathrm{amu}$. Calculate its nuclear density.
Solution:
Here $\mathrm{M}_{\mathrm{Fe}}=55.85 \mathrm{amu}=55.85 \times 1.66 \times 10^{-27} \mathrm{~kg}$ $=9.27 \times 10^{-26} \mathrm{~kg}$
Nuclear Mass $=\mathrm{R}_0 \mathrm{~A}^{1 / 3}=1.1 \times 10^{-15} \times(56)^{1 / 3} \mathrm{~m}$
$
\begin{aligned}
\rho_{\mathrm{nu}}=\frac{\text { Nuclear Mass }}{\text { Nuclear Volume }} & =\frac{\mathrm{M}_{\mathrm{Fe}}}{\frac{4}{3} \pi \mathrm{R}^3} \\
& =\frac{9.27 \times 10^{-26}}{\frac{4}{3} \times 3.14 \times\left(1.1 \times 10^{-15}\right)^3 \times 56} \\
& =\frac{9.27 \times 10^{-26}}{283.688 \times 10^{-45} \times 1.1}=\frac{9.27 \times 10^{+19}}{312.057}
\end{aligned}
$
$\rho_{\mathrm{nu}}=2.9 \times 10^{17} \mathrm{~kg} \mathrm{~m}^{-3}$.
Question 9.
Calculate the density of hydrogen nuclear in SI units. Given $\mathrm{R}_0=1.1$ fermi and $\mathrm{m}_{\mathrm{p}}=$ $1.007825 \mathrm{amu}$.
Solution:
$
\begin{aligned}
& \rho=\frac{3 m_p}{4 \pi R_0^3}=\frac{3 \times 1.007825 \times 1.66 \times 10^{-27}}{4 \times 3.14 \times\left(1.1 \times 10^{-15}\right)^3} \\
&=\frac{5.0189685 \times 10^{-27}}{16.71736 \times 10^{-45}} \\
& \rho=2.98 \times 10^{17} \mathrm{~kg} \mathrm{~m}^{-3} .
\end{aligned}
$

Question 10.
Express one atomic mass unit in energy units, first in Joules and then in $\mathrm{MeV}$. Using this, express the mass defect of ${ }_8^{16} \mathrm{O}$ in $\mathrm{MeV}$.
Solution:
We have, $\mathrm{m}=1 \mathrm{amu}=1.66 \times 10^{-27} \mathrm{~kg}, \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}$
$
\begin{aligned}
& \mathrm{E}=\mathrm{mc}_2=1.66 \times 10^{-27} \times\left(3 \times 10^8\right)^2 \\
& =14.94 \times 10^{-11} \mathrm{~J} \\
& =\frac{1.494 \times 10^{-10}}{1.6 \times 10^{-13}} \mathrm{MeV}\left[1 \mathrm{MeV}=1.6 \times 10^{-13}\right] \\
& =931.5 \mathrm{MeV}
\end{aligned}
$
The ${ }_8^{16} O$ nucleus contains 8 protons and 8 neutrons
Mass of 8 protons $=8 \times 1.00727=8.05816 \mathrm{amu}$
Mass of 8 neutrons $=8 \times 1.00866=\underline{8.06928 \mathrm{amu}}$
Total Mass $=16.12744 \mathrm{amu}$
Mass of ${ }_8^{16} O$ nucleus $=15.99053 \mathrm{amu}$
Mass defect $=\underline{0.13691 \mathrm{amu}}$
$\Delta \mathrm{E}_{\mathrm{b}}=0.13691 \times 931.5 \mathrm{Mev}$
$\Delta \mathrm{E}_{\mathrm{b}}=127.5 \mathrm{Mev}$
Question 11.
The decay constant, for a given redioactive sample is 0.3465 / day. What percentage of this sample will get decayed in a period of 4 years?
Solution:
Here $\lambda,=0.3465 /$ day; $t=4$ years
$
\begin{aligned}
\mathrm{T}_{1 / 2} & =\frac{0.6931}{\lambda}=\frac{0.6931}{0.3465}=2 \text { days } \\
\mathrm{n} & =\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{4}{2}=2
\end{aligned}
$
Hence sample left undecayed after a period of 4 years,
$
\frac{\mathrm{N}}{\mathrm{N}_0}=\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{1}{2}\right)^2=\frac{1}{4}=25 \%
$

Question 12.
If $200 \mathrm{MeV}$ energy is released in the fission of a single nucleus of ${ }_{92}^{235} U$, how many fissions must occur to produce a power of $1 \mathrm{~kW}$ ?
Solution:
Let the number of fissions per second be $\mathrm{n}$.
Then, Energy released per second $=\mathrm{n} \times 200 \mathrm{MeV}$
$=\mathrm{n} \times 200 \times 1.6 \times 10^{-13} \mathrm{~J}$
Energy required per second $=$ Power $x$ Time
$
=1 \mathrm{~kW} \times 1 \mathrm{~s}=1000 \mathrm{~J}
$
Energy released = Energy required
$
\begin{aligned}
& \mathrm{n} \times 200 \times 1.6 \times 10^{-13}=1000 \\
& \mathrm{n}=3.125 \times 10^{-13}
\end{aligned}
$The circuit used to feedback a portion of the output to the input is called the feedback network. If the portion of the output fed to the input is in phase with the input, then the magnitude of the input signal increases. It is necessary for sustained oscillations.