Energy of a photon of frequency $$\nu $$ is given by $$E = h\nu .$$
$$\eqalign{
& {\text{Also,}}\,E = m{c^2},m{c^2} = h\nu \cr
& \Rightarrow mc = \frac{{h\nu }}{C} \Rightarrow p = \frac{{h\nu }}{c} \cr} $$
2.
A telephonic communication service is working at carrier
frequency of $$10\,GHz.$$ Only $$10\% $$ of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of $$5\,kHz$$ ?
If $$n = $$ no. of channels
$$\eqalign{
& 10\% \,{\text{of}}\,10\,GHz = n \times 5\,KHz \cr
& {\text{or,}}\,\frac{{10}}{{100}} \times 10 \times {10^9} = n \times 5 \times {10^3} \Rightarrow n = 2 \times {10^5} \cr} $$
3.
In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength $$\left( \lambda \right)$$ of incident light and the corresponding stopping potential $$\left( {{V_0}} \right)$$ are given below :
$$\lambda \left( {\mu m} \right)$$
$${V_0}$$ (Volt)
0.3
2.0
0.4
1.0
0.5
0.4
Given that $$c = 3 \times {10^8}m{s^{ - 1}}$$ and $$e = 1.6 \times {10^{ - 19}}C,$$ Planck's constant (in units of $$Js$$ ) found from such an experiment is
$$\eqalign{
& \frac{{hc}}{{e{\lambda _1}}} - \frac{\phi }{e} = {V_0}_{_1}\,{\text{and}}\,\frac{{hc}}{{e{\lambda _2}}} - \frac{\phi }{e} = {V_0}_{_2} \cr
& \therefore \frac{{hc}}{e}\left[ {\frac{1}{{{\lambda _1}}} - \frac{1}{{{\lambda _2}}}} \right] = {V_0}_{_1} - {V_0}_{_2} \cr
& \therefore h = \frac{{e\left( {{V_0}_{_1} - {V_0}_{_2}} \right){\lambda _1}{\lambda _2}}}{{\left( {{\lambda _2} - {\lambda _1}} \right)c}} \cr} $$
From the first two values given in data
$$\eqalign{
& h = \frac{{1.6 \times {{10}^{ - 19}}\left[ {2 - 1} \right] \times 0.4 \times 0.3 \times {{10}^{ - 6}}}}{{0.1 \times 3 \times {{10}^8}}} \cr
& h = 0.64 \times {10^{ - 33}} = 6.4 \times {10^{ - 34}}J - s \cr} $$
Similarly if we calculate $$h$$ for the last two values of data $$h = 6.4 \times {10^{ - 34}}J - s$$
4.
The surface of a metal is illuminted with the light of $$400\,nm.$$ The kinetic energy of the ejected photo electrons was found to be $$1.68\,eV.$$ The work function of the metal is :
$$\left( {hc = 1240\,eV.nm} \right)$$
$${E_0} = C{B_0}{\text{ and }}C = \frac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$$
Electric energy density $$ = \frac{1}{2}{\varepsilon _0}E_0^2 = {\mu _E}$$
Magnetic energy density $$ = \frac{1}{2}\frac{{B{o^2}}}{{{\mu _0}}} = {\mu _B}$$
Thus, $${\mu _E} = {\mu _B}$$
Energy is equally divided between electric and magnetic field
6.
Radiation of wavelength $$\lambda ,$$ is incident on a photocell. The fastest emitted electron has speed $$\nu .$$ If the wavelength is changed to $$\frac{{3\lambda }}{4},$$ the speed of the fastest emitted electron will be:
A.
$$ = \nu {\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
B.
$$ = \nu {\left( {\frac{3}{4}} \right)^{\frac{1}{2}}}$$
C.
$$ > \nu {\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
D.
$$ < \nu {\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
7.
The anode voltage of a photocell is kept fixed. The wavelength $$\lambda $$ of the light falling on the cathode is gradually changed. The plate current $$I$$ of the photocell varies as follows :
As $$\lambda $$ is increased, there will be a value of $$\lambda $$ above which photoelectrons will be cease to come out so photocurrent will become zero. Hence (D) is correct answer.
8.
$${K_\alpha }$$ wavelength emitted by an atom of atomic number $$Z = 11$$ is $$\lambda .$$ Find the atomic number for an atom that emits $${K_\alpha }$$ radiation with wavelength $$4\lambda .$$
9.
The work function of a substance is $$4.0\,eV.$$ The longest wavelength of light that can cause photoelectron emission from this substance is approximately.
10.
This question has Statement - 1 and Statement - 2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement - 1: A metallic surface is irradiated by a monochromatic light of frequency $$v > {v_0}$$ (the threshold frequency). The maximum kinetic energy and the stopping potential are $${K_{\max }}$$ and $${V_0}$$ respectively. If the frequency
incident on the surface is doubled, both the $${K_{\max }}$$ and $${V_0}$$ are also doubled. Statement - 2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.
A.
Statement-1 is true, Statement-2 is true, Statement -2 is the correct explanation of Statement - 1.
B.
Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of Statement-1.
C.
Statement-1 is false, Statement-2 is true.
D.
Statement-1 is true, Statement-2 is false.
Answer :
Statement-1 is false, Statement-2 is true.
By Einstein photoelectric equation,
$${K_{\max }} = e{V_0} = hv - h{v_0}$$
When $$v$$ is doubled, $${K_{\max }}$$ and $${V_0}$$ become more than double.