Numerical problems-1 - Chapter 7 Dual Nature of Radiation and Matter 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numericalproblems
Question 1.
How many photons per second emanate from a $50 \mathrm{~mW}$ laser of $640 \mathrm{~nm}$ ?
Answer:
$
\begin{aligned}
& \mathrm{P}=50 \mathrm{~mW} \\
& \lambda=640 \mathrm{~nm} \\
& \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js} \\
& \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}
\end{aligned}
$
Number of photons emanate per second $\mathrm{n}_{\mathrm{p}}=\frac{P}{E}=\frac{P \lambda}{h c}$
$
\begin{aligned}
& =\frac{50 \times 10^3 \times 640 \times 10^{-9}}{6.6 \times 10^{-34} 3 \times 10^8}=\frac{32000 \times 10^{-6}}{19.8 \times 10^{-26}}=1616.16 \times 10^{-6} \\
& \mathrm{n}_{\mathrm{p}}=1.61 \times 1010^{17} \mathrm{~s}^{-1}
\end{aligned}
$
Question 2.
Calculate the maximum kinetic energy and maximum velocity of the photoelectrons emitted when the stopping potential is $81 \mathrm{~V}$ for the photoelectric emission experiment.
Answer:
$
\begin{aligned}
& \mathrm{V}_0=81 \mathrm{~V} \\
& \mathrm{e}=1.6 \times 10^{-19} \mathrm{C} \\
& \mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}
\end{aligned}
$
Maximum kinetic energy of electron,
$
\begin{aligned}
& \mathrm{K}_{\max }=\mathrm{e}^{\mathrm{V}_{\mathrm{o}}} \\
& =1.6 \times 10^{-19} \times 81 \\
& =129.6 \times 10^{-19} \\
& =1.29 \times 10^{-17} \\
& \mathrm{~K}_{\max }=1.3 \times 10^{-17} \mathrm{~J}
\end{aligned}
$
aximum velocity of photoelectron,
$
v_{\max }=\sqrt{\frac{2 e \mathrm{~V}_0}{m}}
$

$
\begin{aligned}
& =\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 81}{9.1 \times 10^{-31}}}=\sqrt{\frac{259.2 \times 10^{-19}}{9.1 \times 10^{-31}}}=\sqrt{28.48 \times 10^{12}} \\
v_{\max } & =5.3 \times 10^6 \mathrm{~ms}^{-1}
\end{aligned}
$
Question 3.
Calculate the energies of the photons associated with the following radiation:
(i) violet light of $413 \mathrm{~nm}$
(ii) X-rays of $0.1 \mathrm{~nm}$
(iii) radio waves of $10 \mathrm{~m}$.
Answer:
$
\begin{aligned}
& \mathrm{h}=6.6 \times 10^{-34} \mathrm{Js} \\
& \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}
\end{aligned}
$
Energy of photon, $\mathrm{E}=\mathrm{h} v$
$
\mathrm{E}=\frac{h c}{\lambda}
$
(i) Violet light, $\lambda=413 \mathrm{~nm}$
$
\begin{aligned}
\mathrm{E} & =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{413 \times 10^{-9}}=0.04794 \times 10^{-17} \\
& =4.794 \times 10^{-19} \mathrm{~J} \\
& =\frac{4.794 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV} \\
E & =3 \mathrm{eV}
\end{aligned}
$

(ii) X-rays of, $\lambda=0.1 \mathrm{~nm}$
$
\begin{aligned}
\mathrm{E} & =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{0.1 \times 10^{-9}}=198 \times 10^{-17} \mathrm{~J} \\
& =\frac{198 \times 10^{-17}}{1.6 \times 10^{-19}}=123.75 \times 10^2 \\
\mathrm{E} & =12375 \mathrm{eV}
\end{aligned}
$
(iii) Radio waves, $\lambda=10 \mathrm{~m}$
$
\begin{aligned}
E & =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{10}=1.98 \times 10^{-26} \mathrm{~J} \\
& =\frac{1.98 \times 10^{-26}}{1.6 \times 10^{-19}}=1.2375 \times 10^{-7} \\
E & =1.24 \times 10^{-7} \mathrm{eV}
\end{aligned}
$

Question 4.
A $150 \mathrm{~W}$ lamp emits light of mean wavelength of $5500 \AA$. If the efficiency is $12 \%$, find out the number of photons emitted by the lamp in one second.
Answer:
$
\begin{aligned}
& \mathrm{P}=150 \mathrm{~W} \\
& \lambda=5500 \AA \\
& \mathrm{h}=6.6 \times 10^{-34} \mathrm{Js} \\
& \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}
\end{aligned}
$
Number of photons emitted per second $\mathrm{n}=\frac{p \lambda}{h c}$
If the efficiency is $12 \%, \eta=\frac{12}{100}=0.12$
$
\begin{aligned}
\mathrm{n} & =\frac{p \eta \lambda}{h c} \\
& =\frac{150 \times 0.12 \times 5500 \times 10^{-10}}{6.6 \times 10^{-34} \times 3 \times 10^8}=\frac{99000 \times 10^{-10}}{19.8 \times 10^{-26}}=5000 \times 10^{16} \\
\mathrm{n} & =5 \times 10^{19}
\end{aligned}
$
Question 5.
How many photons of frequency $10^{14} \mathrm{~Hz}$ will make up $19.86 \mathrm{~J}$ of energy?
Answer:
Total energy emitted per second $=$ Power $\mathrm{x}$ time
$19.86 \mathrm{~J}=$ Power $\mathrm{x}$ is
$\therefore$ Power $=19.86 \mathrm{~W}$
Number of photons, $\mathrm{n}=\frac{p}{E}=\frac{p}{h v}$
$
\begin{aligned}
& =\frac{19.86}{6.6 \times 10^{-34} \times 10^{14}}=3.009 \times 10^{20} \\
& \mathrm{n}=3 \times 10^{20} \\
& \mathrm{n}_{\mathrm{p}}=3 \times 10^{20}
\end{aligned}
$

Question 6.
What should be the velocity of the electron so that its momentum equals that of $4000 \AA$ wavelength photon.
Answer:
de-Broglie wavelength of electron
$
\begin{aligned}
& \lambda=\frac{h}{p} \\
& \mathrm{v}=\frac{h}{m \lambda} \\
& =\frac{6.6 \times 10^{-34}}{9.11 \times 10^{-31} \times 4000 \times 10^{-10}}=\frac{6.6 \times 10^{-34}}{36.44 \times 10^{-38}}=0.18112 \times 10^4 \\
& \mathrm{v}=1811 \mathrm{~ms}^{-1}
\end{aligned}
$
Question 7.
When a light of frequency $9 \times 10^{14} \mathrm{~Hz}$ is incident on a metal surface, photoelectrons are emitted with a maximum speed of $8 \times 10^5 \mathrm{~ms}^{-1}$. Determine the threshold frequency of the surface.
Answer:
According to Einstein's photoelectric equation
$
\begin{aligned}
& \frac{1}{2} m v_{\max }^2=\mathrm{h}\left(v_{\left.-v_0\right)}\right. \\
& \begin{aligned}
v_0 & =v^{-}-\frac{m v_{\max }^2}{2 h} \\
& =9 \times 10^{14}-\left(\frac{9.11 \times 10^{-31} \times\left(8 \times 10^5\right)^2}{2 \times 6.6 \times 10^{-34}}\right) \\
& =9 \times 10^{14}-\left(\frac{583.04 \times 10^{-21}}{13.2 \times 10^{-34}}\right)=9 \times 10^{14}-44.169 \times 10^{13}
\end{aligned} \\
& =9 \times 0^{14}-4.4 \times 10^{14} \\
& v_0=4.6 \times 10^{14} \mathrm{~Hz}
\end{aligned}
$

Question 8.
When a $6000 \AA$ light falls on the cathode of a photo cell and produced photoemission. If a stopping potential of $0.8 \mathrm{~V}$ is required to stop emission of electron, then determine the:
1. frequency of the light
2. energy of the incident photon
3. work function of the cathode material
4. threshold frequency
5. net energy of the electron after it leaves the surface.
Answer:
Wavelength, $\lambda=6000 \AA=6000 \times 10^{-10} \mathrm{~m}$
stopping potential, $\mathrm{V}_0=0.8 \mathrm{~V}$
1. Frequency of the light, $v=\frac{c}{\lambda}$
$=\frac{3 \times 10^8}{600 \times 10^{-10}}=5 \times 10^4 \times 10^{-18}$
$v=5 \times 10^{14} \mathrm{~Hz}$
2. Energy of the incident photon,
$
\begin{aligned}
& E=h v=6.6 \times 10^{-34} \times 5 \times 10^{14} \\
& =33 \times 10^{-20} \mathrm{~J}
\end{aligned}
$

$
\begin{aligned}
& =\frac{33 \times 10^{-20}}{1.6 \times 10^{-19}}=20.625 \times 10^{-1} \\
& E=2.06 \mathrm{eV}
\end{aligned}
$
3. Work function of the cathode material.
$
\begin{aligned}
& \mathrm{W}_0=\mathrm{h} v-\mathrm{eV}_0 \\
& =\left(\frac{6.6 \times 10^{-34} \times 5 \times 10^{14}}{1.6 \times 10^{-19}}\right)-\left(\frac{1.6 \times 10^{-19} \times 0.8}{1.6 \times 10^{-19}}\right)=2.06-0.8 \\
& \mathrm{~W}_0=1.26 \mathrm{eV}
\end{aligned}
$
4. Threshold frequency, $\mathrm{W}_0=h v_0$
$
\begin{aligned}
& v_0=\frac{W_0}{h}=\frac{1.26 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}}=0.3055 \times 10^{15} \\
& v_0=3.05 \times 10^{14} \mathrm{~Hz}
\end{aligned}
$
5. Net energy of the electron after it leaves the surface
$
\begin{aligned}
& E=\left(v-v_0\right) \\
& =6.6 \times 10^{-34}\left(5 \times 10^{14}-3.06 \times 10^{14}\right. \\
& =6.6 \times 10^{-34} \times 1.94 \times 10^{14} \\
& E=12.804 \times 10^{-20} \mathrm{~J} \\
& =\frac{1.2804 \times 10^{-19}}{1.6 \times 10^{-19}} \\
& E=0.8 \mathrm{e} V
\end{aligned}
$

Question 9.
A $3310 \AA$ photon liberates an electron from a material with energy $3 \times 10^{-19} \mathrm{~J}$ while another $5000 \AA$ photon ejects an electron with energy $0.972 \times 10^{-19} \mathrm{~J}$ from the same material. Determine the value of Planck's constant and the threshold wavelength of the material.
Answer:
They energy of ejected electron is given by $\mathrm{E}=\frac{h c}{\lambda}-\frac{h c}{\lambda_0}$
$
\begin{aligned}
& 3 \times 10^{-19}=h c\left[\frac{1}{3310 \times 10^{-10}}-\frac{1}{\lambda_0}\right] \\
& 9.72 \times 10^{-20}=h c\left[\frac{1}{5000 \times 10^{-10}}-\frac{1}{\lambda_0}\right]
\end{aligned}
$
Subtracting (2) from (1), we get
$
\begin{aligned}
& (3-0.972) \times 10^{-19}=\frac{h c}{10^{-10}}\left[\frac{1}{3310}-\frac{1}{5000}\right] \\
& 2.028 \times 10^{-19}=\frac{h \times 3 \times 10^8}{10^{-10}}\left[\frac{1690}{3310 \times 5000}\right] \\
& h=\frac{2.028 \times 10^{-19} \times 10^{-10} \times 3310 \times 5000}{3 \times 10^8 \times 1690} \\
& h=6.62 \times 10^{-34} \mathrm{Js} \\
& \text { Now } \quad \mathrm{W}_0=\frac{h c}{\lambda}-\mathrm{E} \\
&
\end{aligned}
$

$
\begin{aligned}
& =\left(\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{3310 \times 10^{-10}}\right)-3 \times 10^{-19}=(6-3) \times 10^{-19} \\
\mathrm{~W}_0 & =3 \times 10^{-19} \mathrm{~J}
\end{aligned}
$
Threshold Wavelength,
$
\begin{aligned}
\lambda_0 & =\frac{h c}{\mathrm{~W}_0} \\
& =\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{3 \times 10^{-19}}=6.62 \times 10^{-7} \mathrm{~m} \\
\lambda_0 & =6620 \times 10^{-10} \mathrm{~m}
\end{aligned}
$
Question 10.
At the given point of time, the earth receives energy from Sun at $4 \mathrm{cal} \mathrm{cm}^{-2} \mathrm{~min}^{-1}$. Determine the number of photons received on the surface of the Earth per $\mathrm{cm}^2$ per minute. (Given : Mean wavelength of Sun light = $5500 \AA$ )
Answer:
$
\begin{aligned}
& E=4 \text { calorie } \\
& =4 \times 4.184 \mathrm{~J} \\
& \lambda=5500 \AA
\end{aligned}
$
Number of photons received on the surface of the earth, from $E=n h v$
$
\begin{aligned}
& \mathrm{n}=\frac{E \lambda}{h c} \\
& =\frac{4 \times 4.184 \times 5500 \times 10^{-10}}{6.6 \times 10^{-34} \times 3 \times 10^8}=\frac{9.2048 \times 10^{-10}}{19.8 \times 10^{-26}}=4648 \times 10^{16}
\end{aligned}
$
$
\begin{aligned}
& =4.648 \times 10^{19} \\
& \mathrm{n}=4.65 \times 10^{19}
\end{aligned}
$

Question 11.
UV light of wavelength $1800 \AA$ is incident on a lithium surface whose threshold wavelength $4965 \AA$. Determine the maximum energy of the electron emitted.
Answer:
$
\begin{aligned}
& \lambda=1800 \times 10^{-10} \mathrm{~m} \\
& \lambda_0=4965 \times 10^{-10} \mathrm{~m} \\
& \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js} \\
& \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}
\end{aligned}
$
Maximum kinetic energy of electron,
$
\begin{aligned}
\mathrm{K}_{\max } & =h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) \\
& =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{10^{-10}} \times\left(\frac{1}{1800}-\frac{1}{4965}\right) \\
& =19.8 \times 10^{-16} \times \frac{3165}{8937 \times 10^3}=7.01208 \times 10^{-19} \mathrm{~J} \\
& =\frac{7.01208 \times 10^{-19}}{1.6 \times 10^{-19}}=4.38 \mathrm{eV} \\
\mathrm{K}_{\max } & =4.40 \mathrm{eV}
\end{aligned}
$
Question 12.
Calculate the de Broglie wavelength of a proton whose kinetic energy is equal to $81.9 \mathrm{x}$ $10^{-15} \mathrm{~J}$. (Given: mass of proton is 1836 times that of electron).
Answer:
$
\begin{aligned}
& \mathrm{m}_{\mathrm{p}}=1.67 \times 10^{-27} \mathrm{~kg} \\
& \mathrm{~K} . \mathrm{E}=81.9 \times 10^{-15} \mathrm{~J}
\end{aligned}
$
de-Broglie wavelength of proton, $\lambda=\frac{h}{\sqrt{2 m K}}$
$
=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 81.9 \times 10^{-15}}}=\frac{6.6 \times 10^{-34}}{1.6539 \times 10^{-20}}=3.99 \times 10^{-14}
$

$
\lambda=4 \times 10^{-14} \mathrm{~m}
$
Question 13.
A deuteron and an alpha particle are accelerated with the same potential. Which one of the two has (i) greater value of de Broglie wavelength associated with it and (ii) less kinetic energy? Explain.
Answer:
(i) Using de-Broglie wavelength formula, the dueteron and alpha particle are accelerated with same potential. So, both their velocities are same.
$
\lambda=\frac{h}{\sqrt{2 m V_0 q}}, \lambda \propto \frac{1}{\sqrt{m q}}
$
So, $\lambda_{\text {due }} \propto \frac{1}{\sqrt{2 m_d q_d}}$ and $\lambda_\alpha \propto \frac{1}{\sqrt{8 m_\alpha q_\alpha}}$
$
\begin{aligned}
& \frac{\lambda_\alpha}{\lambda_d}=\frac{\sqrt{2 m_d q_d}}{\sqrt{8 m_\alpha q_\alpha}}=\sqrt{\frac{2}{8}}=\sqrt{\frac{1}{4}}=\frac{1}{2} \\
& \lambda_d=2 \lambda_\alpha
\end{aligned}
$
(ii) For same potential of acceleration, $\mathrm{KE}$ is directly proportional to the ' $q$ '
Charge of duetron is $+e$
Charge of alpha is $+2 \mathrm{e}$
So, $\mathrm{K}_{\mathrm{d}}=\frac{K_\alpha}{2}$
Charge of alpha particle is more than the duetron.
Question 14.
An electron is accelerated through a potential difference of $81 \mathrm{~V}$. What is the de Broglie wavelength associated with it? To which part of electromagnetic spectrum does this wavelength correspond?
Answer:

de-Broglie wavelength of an electron beam accelerated through a potential difference of $\mathrm{V}$ volts is
$
\begin{aligned}
& \lambda=\frac{h}{\sqrt{2 m e V}}=\frac{1.23}{\sqrt{V}} \mathrm{~nm} \\
& \mathrm{~V}=81 \mathrm{~V} \text {, so } \lambda=\frac{1.23}{\sqrt{81}} \times 10^{-9} \mathrm{~m} \\
& \lambda=1.36 \AA
\end{aligned}
$
$\mathrm{X}$-ray is the part of electromagnetic spectrum does this wavelength corresponds. X-ray has the wavelengths ranging from about $10^8$ to $10^{-12} \mathrm{~m}$.
Question 15.
The ratio between the de Broglie wavelengths associated with protons, accelerated
through a potential of $512 \mathrm{~V}$ and that of alpha particles accelerated through a potential of $\mathrm{X}$ volts is found to be one. Find the value of $\mathrm{X}$.
Answer:
de-Broglie wavelength of accelerated charge particle
$
\begin{aligned}
& \lambda=\frac{h}{\sqrt{2 m q V}} \\
& \lambda \propto \frac{h}{\sqrt{m q V}}
\end{aligned}
$
Ratio of wavelength of proton and a-particle.
$
\frac{\lambda_p}{\lambda_\alpha}=\sqrt{\frac{m_\alpha q_\alpha \mathrm{V}_\alpha}{m_p q_p \mathrm{~V}_p}}=\sqrt{\left(\frac{m_\alpha}{m_p}\right)\left(\frac{q_\alpha}{q_p}\right)\left(\frac{\mathrm{V}_\alpha}{\mathrm{V}_p}\right)}
$
Here, $\frac{m_\alpha}{m_p}=4 ; \frac{q_\alpha}{q_p}=2 ; \frac{\mathrm{V}_\alpha}{\mathrm{V}_p}=\frac{\mathrm{X}}{512} ; \frac{\lambda_p}{\lambda_\alpha}=1$
$
\begin{aligned}
& 1=\sqrt{4 \times 2 \times\left(\frac{X}{512}\right)}=\sqrt{\frac{X}{64}}=\frac{X}{64} \\
& X=64 \mathrm{~V}
\end{aligned}
$