81.
The ratio of the $${\lambda _{\min }}$$ in a Coolidge tube to $${\lambda _{{\text{de Broglie}}}}$$ of the electrons striking the target depends on accelerating potential $$V$$ as
A
$$\frac{{{\lambda _{\min }}}}{{{\lambda _{{\text{de Broglie}}}}}} \propto \sqrt V $$
B
$$\frac{{{\lambda _{\min }}}}{{{\lambda _{{\text{de Broglie}}}}}} \propto V$$
C
$$\frac{{{\lambda _{\min }}}}{{{\lambda _{{\text{de Broglie}}}}}} \propto \frac{1}{{\sqrt V }}$$
D
$$\frac{{{\lambda _{\min }}}}{{{\lambda _{{\text{de Broglie}}}}}} \propto \frac{1}{V}.$$
82.
The $$21\,cm$$ radiowave emitted by hydrogen in interstelar space is due to the interaction called the hyperfine interaction in atomic hydrogen. The energy of the emitted wave is nearly
The energy of emitted photon is given by
$$E = \frac{{hc}}{\lambda }$$
Given, wavelength, $$\lambda = 21\,cm = 0.21\,m$$
So, $$E = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{0.21}}$$
$$ = {10^{ - 24}}\,J$$
83.
$$1.5\,mW$$ of $$400\,nm$$ light is directed at a photoelectric cell. If 0.10 per cent of the incident photons produce photoelectrons, then find the current in the cell.
The number of photons per second directed at the cell is
$$\eqalign{
& n = \frac{{{\text{Power}}}}{{hv}} = \frac{{P\lambda }}{{hc}} = \frac{{1.5 \times {{10}^{ - 3}} \times 400 \times {{10}^{ - 9}}}}{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} \cr
& n = \frac{1}{{3.31}} \times {10^{16}}\left( {{\text{photons/s}}} \right) \cr} $$
But it is given that only $$0.1\% $$ of these photons produce photoelectrons. Therefore number of photoelectrons produced per second is
$${n_e} = \frac{{0.1}}{{100}} \times \frac{{{{10}^{16}}}}{{3.31}} = \frac{{{{10}^{13}}}}{{3.31}}$$
Therefore, the current is
$$\eqalign{
& I = {n_e}e \cr
& = \frac{{{{10}^{13}} \times 1.6 \times {{10}^{ - 19}}}}{{3.31}} = 0.48 \times {10^{ - 6}} = 0.48\,\mu A \cr} $$
84.
An electron of mass $$m$$ and a photon have same energy $$E.$$ The ratio of de-Broglie wavelengths associated with them is
($$c$$ being velocity of light)
A
$${\left( {\frac{E}{{2m}}} \right)^{\frac{1}{2}}}$$
B
$$c{\left( {2mE} \right)^{\frac{1}{2}}}$$
C
$$\frac{1}{c}{\left( {\frac{{2m}}{E}} \right)^{\frac{1}{2}}}$$
D
$$\frac{1}{c}{\left( {\frac{E}{{2m}}} \right)^{\frac{1}{2}}}$$
Since, it is given that electron has mass $$m.$$ de-Broglie’s wavelength for an electron will be given as
$${\lambda _e} = \frac{h}{P}\,......\left( {\text{i}} \right)$$
where, $$h$$ = Planck’s constant
$$P$$ = Linear momentum of electron
As kinetic energy of electron
$$E = \frac{{{P^2}}}{{2m}} \Rightarrow P = \sqrt {2mE} \,.......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we get
$${\lambda _e} = \frac{h}{{\sqrt {2mE} }}\,......\left( {{\text{iii}}} \right)$$
Energy of a photon can be given as
$$\eqalign{
& E = hv \cr
& \Rightarrow E = \frac{{hc}}{{{\lambda _p}}} \cr
& \Rightarrow {\lambda _p} = \frac{{hc}}{E}\,......\left( {{\text{iv}}} \right) \cr} $$
Hence, $${\lambda _p} = $$ de-Broglie's wavelength of photon. Now, divide Eq. (iii) by Eq. (iv), we get
$$\eqalign{
& \frac{{{\lambda _e}}}{{{\lambda _p}}} = \frac{h}{{\sqrt {2mE} }} \cdot \frac{E}{{hc}} \cr
& \Rightarrow \frac{{{\lambda _c}}}{{{\lambda _p}}} = \frac{1}{c} \cdot \sqrt {\frac{E}{{2m}}} \cr} $$
85.
A source $${S_1}$$ is producing, $${10^{15}}$$ photons/s of wavelength $$5000\,\mathop {\text{A}}\limits^ \circ .$$ Another source $${S_2}$$ is producing $$1.02 \times {10^{15}}$$ photons per second of wavelength $$5100\,\mathop {\text{A}}\limits^ \circ .$$ Then, $$\frac{{\left( {{\text{power}}\,{\text{of}}\,{S_2}} \right)}}{{\left( {{\text{power}}\,{\text{of}}\,{S_1}} \right)}}$$ is equal to
Number of photons emitted per second is given by
$$\eqalign{
& n = \frac{P}{{\left( {\frac{{hc}}{\lambda }} \right)}}\,\,\,\left[ {_{\frac{{hc}}{\lambda }\, = \,{\text{Energy}}}^{P\, = \,{\text{Power}}}} \right] \cr
& {\text{So,}}\,\,P = \frac{{nhc}}{\lambda } \cr} $$
So, for two different situations,
$$ \Rightarrow \frac{{{P_2}}}{{{P_1}}} = \frac{{{n_2}{\lambda _1}}}{{{n_1}{\lambda _2}}} = \frac{{1.02 \times {{10}^{15}} \times 5000}}{{{{10}^{15}} \times 5100}} = 1$$
86.
An X-ray tube is operated at $$15\,kV.$$ Calculate the upper limit of the speed of the electrons striking the target.
The maximum kinetic energy of an electron accelerated through a potential difference of $$V$$ volt is $$\frac{1}{2}m{v^2} = eV$$
∴ maximum velocity $$v = \sqrt {\frac{{2eV}}{m}} $$
$$\eqalign{
& v = \sqrt {\frac{{2 \times 1.6 \times {{10}^{ - 19}} \times 15000}}{{9.1 \times {{10}^{ - 31}}}}} , \cr
& v = 7.26 \times {10^7}\,m/s \cr} $$
87.
The threshold wavelength of the tungsten is $$2300\,\mathop {\text{A}}\limits^ \circ .$$ If ultraviolet light of wavelength $$1800\,\mathop {\text{A}}\limits^ \circ $$ is incident on it, then the maximum kinetic energy of photoelectrons would be about -
88.
If the momentum of electron is changed by $$P,$$ then the de Broglie wavelength associated with it changes by $$0.5\% .$$ The initial momentum of electron will be:
The de-Broglie’s wavelength associated with the moving electron $$\lambda = \frac{h}{P}$$
Now, according to problem
$$\eqalign{
& \frac{{d\lambda }}{\lambda } = - \frac{{dp}}{P}; \cr
& \frac{{0.5}}{{100}} = \frac{P}{{P'}} \cr
& \Rightarrow P' = 200\,P \cr} $$
89.
According to Einstein’s photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal Versus the frequency, of the incident radiation gives a straight line whose slope
A
depends both on the intensity of the radiation and the metal used
B
depends on the intensity of the radiation
C
depends on the nature of the metal used
D
is the same for the all metals and independent of the intensity of the radiation
Answer :
is the same for the all metals and independent of the intensity of the radiation
From Equation $$K.E. = hv - \phi $$
slope of graph of $$K.E\,\& \,v$$ is $$h$$ (Plank's constant)
which is same for all metals
90.
The short wavelength limit of continuous X-radiation emitted by an X-ray tube operating at $$30\,kV$$ is $$0.414\,\mathop {\text{A}}\limits^ \circ .$$ Calculate Planck's constant.