Numerical Problems-2 - Chapter 7 Dual Nature of Radiation and Matter 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
If a light of wavelength $4950 \AA$ is viewed as a continuous flow of photons, what is the energy of each photon in $\mathrm{eV}$ ? (Given $\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}$ )
Solution:
Here $\lambda=4950 \AA=4950 \times 10^{-10} \mathrm{~m}$
Energy of each photon,
$
\begin{aligned}
& \mathrm{E}=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{4950 \times 10^{-10}}=4 \times 10^{-19} \mathrm{~J} \\
& =\frac{4 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV} \\
& \mathrm{E}=2.5 \mathrm{eV}
\end{aligned}
$
Question 2.
Monochromatic light of frequency $6 \times 10^{14} \mathrm{~Hz}$ is produced by a laser. The power emitted is $2 \times 10^{-3} \mathrm{w}$.
(i) What is the energy of each photon in the light?
(ii) How many photons per second, on the average, are emitted by the source?
Solution:
(i) Energy of each photon,
$
\begin{aligned}
& E=h v=6.6 \times 10^{-34} \times 6 \times 10^{14} \\
& E=3.98 \times 10^{-19} \mathrm{~J}
\end{aligned}
$
(ii) If $\mathrm{N}$ is the number of photons emitted per second by the source, then Power transmitted in the beam $=\mathrm{N} \mathrm{x}$ energy of each photon
$
\begin{aligned}
& \mathrm{P}=\mathrm{N} \\
& \mathrm{N}=\frac{P}{E}=\frac{2 \times 10^{-3}}{3.98 \times 10^{-19}} \\
& \mathrm{~N}=5 \times 10^{15} \text { Photons per second. }
\end{aligned}
$

Question 3.
Light of wavelength $5000 \AA$ falls on a metal surface of work function $1.9 \mathrm{eV}$. Find:
(i) the energy of photons in eV
(ii) the K.E of photoelectrons and
(iii) the stopping potential.
Solution:
Here $\lambda=5000 \AA=5 \times 10^{-7} \mathrm{~m}$
$\mathrm{W}_0=1.9 \mathrm{ev}$
(i) Energy of a photon,
$
\begin{aligned}
& \mathrm{E}=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{5 \times 10^{-7}} \mathrm{~J}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{5 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{eV} \mathrm{eV} \\
& \mathrm{E}=2.475 \mathrm{eV}
\end{aligned}
$
(ii) K.E of a photoelectron,
K.E $=\mathrm{hv}-\mathrm{W}_0=2.475-1.9=0.575 \mathrm{eV}$
(iii) Let $\mathrm{V}_0$ be the stopping potential. Then
$
\begin{aligned}
& \mathrm{eV}_0=\frac{1}{2} \mathrm{mv}^2=\mathrm{K} . \text { E of a photoelectron } \\
& \mathrm{V}_0=\frac{0.575}{e} \mathrm{eV} \\
& \mathrm{V}_0=0.575 \mathrm{~V}
\end{aligned}
$

Question 4.
If photoelectrons are to be emitted from a potassium surface with a speed $6 \times 10^6 \mathrm{~ms}^{-1}$, what frequency of radiation must be used? (Threshold frequency for potassium is $4.22 \mathrm{x}$ $\left.10^{14} \mathrm{~Hz}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{me}=9.1 \times 10^{-31} \mathrm{~kg}\right)$
Solution:
Here, $\mathrm{v}=6 \times 10^6 \mathrm{~ms}^{-1}$
$
\mathrm{V}_0=4.22 \times 10^{14} \mathrm{~Hz}
$
From Einstein's photoelectric equation,
$
\begin{aligned}
& \mathrm{k} . \mathrm{E}=\frac{1}{2} \mathrm{mv}^2=\mathrm{h}\left(\mathrm{v}-\mathrm{v}_0\right) \\
& v=\frac{1}{2} \frac{m v^2}{h}+v_0 \\
& =\frac{1}{2} \times \frac{9.1 \times 10^{-31}+\left(6 \times 10^6\right)^2}{6.6 \times 10^{-34}}+4.22 \times 10^{-14} \\
& =\left(2.48 \times 10^{14}\right)+\left(4.22 \times 10^{14}\right) \\
& v=6.7 \times 10^{14} \mathrm{~Hz}
\end{aligned}
$
Question 5.
The photoelectric cut-off voltage in a certain experiment $1.5 \mathrm{~V}$. What is the maximum kinetic energy of photoelectrons emitted?
Solution:
Here $\mathrm{V}_0=1.5 \mathrm{~V}$
$\mathrm{K}_{\max }=\mathrm{eV}_0=1.5 \mathrm{eV}$
$=1.5 \times 1.6 \times 10^{-19} \mathrm{~J}$
$\mathrm{K}_{\max }=24 \times 10^{-19} \mathrm{~J}$

Question 6.
What is the (a) momentum, (b) speed, and (c) de-Broglie wavelength of an electron with kinetic energy of $120 \mathrm{eV}$.
Solution:
Kinetic energy, K.E $=120 \mathrm{eV}=120 \times 1.6 \times 10^{-19}$
$\mathrm{K}=\mathrm{K} . \mathrm{E}=1.92 \times 10^{-17} \mathrm{~J}$
(a) Momentum of an electron, $\mathrm{P}=\sqrt{2 m K}$
$\mathrm{P}=\sqrt{2 \times 9.1 \times 10^{-31} \times 1.92 \times 10^{-17}}$
$\mathrm{P}=5.91 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}$
(b) Speed of an electron,
$\mathrm{v}=\frac{p}{m}=\frac{5.91 \times 10^{-24}}{9.1 \times 10^{-31}}=6.5 \times 10^6 \mathrm{~kg} \mathrm{~ms}^{-1}$
(c) de-Broglie wavelength,
$\lambda=\frac{h}{p}=\frac{6.6 \times 10^{-34}}{5.910^{-24}}=1.117 \times 10^{-10}=0.112 \times 10^{-9} \mathrm{~m}$
$\lambda=0.112 \mathrm{~nm}$

Question 7.
An electron and a photon each have a wavelength of $1 \mathrm{~nm}$. Find, (a) their momenta (b) the energy of the photon, and (c) kinetic energy of electron.
Solution:
(a) Both electron and photon have same wavelength. so, they have same momentum also, $\mathrm{P}=\frac{h}{\lambda}=\frac{6.6 \times 10^{-34}}{1 \times 10^{-9}}=6.6 \times 10^{-25} \mathrm{~kg} \mathrm{~ms}^{-1}$
(b) Energy of a photon,
$
\begin{aligned}
\mathrm{E}=\frac{h c}{\lambda} & =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1 \times 10^{-9}}=19.8 \times 10^{-17} \mathrm{~J} \\
& =\frac{19.8 \times 10^{-17}}{1.6 \times 10^{-19}}=12.375 \times 10^2=1.24 \times 10^3 \mathrm{eV} \\
\mathrm{E} & =1.24 \mathrm{keV}
\end{aligned}
$
(c) Kinetic energy of electron,
$
\begin{aligned}
& \mathrm{K}=\frac{p^2}{2 m} \\
& =\frac{\left(6.6 \times 10^{-25}\right)^2}{2 \times 9.1 \times 10^{-31}}=\frac{43.56 \times 10^{-50}}{18.2 \times 10^{-31}}=2.39 \times 10^{-19} \mathrm{~J} \\
& =\frac{2.39 \times 10^{-19}}{1.6 \times 10^{-19}} \\
& \mathrm{~K}=1.49 \mathrm{eV}
\end{aligned}
$
Question 8.
Find the ratio of de-broglie wavelengths associated with two electron beams accelerated through $25 \mathrm{~V}$ and $36 \mathrm{~V}$ respectively.
Solution:
de-Broglie wavelength associate with potential difference $\lambda \propto \frac{1}{\sqrt{ } V}$
$
\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{V_2}{V_1}}=\sqrt{\frac{36}{25}}=\frac{6}{5} \Rightarrow \lambda_1: \lambda_2=6: 5
$

Question 9.
A proton and an alpha particle, both initially at rest, are accelerated so as to have the same kinetic energy. What is the ratio of their de-Broglie wavelength?
Solution:
de-Broglie wavelength,
$
\begin{aligned}
& \lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m K}} \\
& \text { i.e. } \\
& \lambda \propto \frac{1}{\sqrt{ } m}\left[\mathrm{~m}_\alpha=4 \mathrm{~m}_{\mathrm{p}}\right] \\
& \frac{\lambda_p}{\lambda_\alpha}=\sqrt{\frac{m_\alpha}{m_p}}=\sqrt{\frac{4 m_p}{m_p}}=\sqrt{\frac{4}{1}}=\frac{2}{1} \\
& \lambda_{\mathrm{p}}: \lambda_\alpha=2: 1
\end{aligned}
$
Question 10.
Light of two different frequencies whose photons have energies $1 \mathrm{eV}$ and $2.5 \mathrm{eV}$ respectively illuminate a metallic surface whose work function is $0.5 \mathrm{eV}$ successively. Find the ratio of maximum speeds of emitted electrons.
Solution:
$
\mathrm{K}_{\max }=\frac{1}{2} m v_{\max }^2=h \mathrm{v}-\mathrm{W}_0
$
$
\frac{v_{\max }^2(1)}{v_{\max }^2(2)}=\frac{(1-0.5) e \mathrm{~V}}{(2.5-0.5) e \mathrm{~V}}=\frac{0.5}{2}=\frac{1}{4}
$
$
\frac{v_{\max }(1)}{v_{\max }(2)}=\frac{1}{2}=1: 2
$