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97 Questions around this concept.
When photons of wavelength λ1 are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength λ2 are used, the corresponding stopping potential is thrice that of the above value. If light of wavelength λ3 is used then find the stopping potential for this case :
Question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement-1: When ultraviolet light is incident on a photocell, its stopping potential is and the maximum kinetic energy of the photoelectrons is
. When the ultraviolet light is replaced by X-rays, both
and
increase.
Statement-2: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.
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 (the threshold frequency). The maximum kinetic energy and the stopping potential are
and
respectively. If the frequency incident on the surface is doubled, both the
and
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.
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Two identical photo cathodes receive light of frequencies . If the velocities of the photoelectrons (of mass m ) coming out are respectively
, then
According to Einstein’s photoelectric equation, the plot of the kinetic energy of the emitted photoelectrons from a metal and the frequency of the incident radiation gives a straight line whose slope is:
Photons of energies $1 eV$ and $2 eV$ are successively incident on a metallic surface of work function $0.5 eV$. The ratio of kinetic energy of most energetic photoelectrons in the two cases will be.
The electric field of light wave is given as $\vec{E}=10^{-6} \cos \left(\frac{2 \pi x}{5 \times 10^{-7}}-7 \pi \times 6 \times 10^{14} t\right) x \mathrm{N} / \mathrm{c}$.This light fall on metal plate of work function, if the stopping potential of the photoelectrons is 0.48 v.
Given $t($ in ev $)=\frac{12375}{\lambda(\text { inA })}$
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The stopping potential for photoelectric current is found to be $2 \nu_0$ when we incident light of wavelength $\lambda$ on a metallic surface. But if we increased the wavelength of light by $50 \%$. Then the stopping potential was found to be $\nu_0$ what is the value of the threshold wavelength for the surface?
The work function of metal A and B are in ratio 3:2 then what is the ratio of their corresponding threshold wavelength?
Choose the correct relation according to Einstein's photoelectric equation:
Einstein's Photoelectric equation-
As we studied that E = hν is the equation of energy of each photon. Now we have also studied that the threshold frequency is that frequency below which the electrons won’t come out of the metallic surface. From the above equation we see that Energy is a function of frequency. Now one question will come in mind that when the electron gets ejected then where does all the electron goes?? Does that electron have some energy to go any where??? If yes then which type of energy??
All these type of question was well answered by the greta scientist Albert Einstein, According to the experiment performed by the Albert Einstein, there are some conclusion that those electron have kinetic energy only. Also the energy absorbed by the photons is partly used to overcome the force by the metallic surface. SInce there is no electric field present outside the metallic surface so there will be only energy present is pure kinetic energy.
So, we have K.E. of the photo-electrons = (Energy obtained from the Photon) – (The energy used to escape the metallic surface)
Here, The energy used to escape the metallic surface is the wrok function $(\phi)$ which we have discussed already. So the Einstein's Photoelectric equation can also be written as -
$$
\text { K.E. }=\mathrm{hv}-\Phi
$$
We can understand the work function more clearly like this -
As we know that an electron needs some minimum energy to be extracted from a metallic surface. So from the above equation, if $\mathrm{v}=$ threshold frequency ( $\mathrm{v}_0$ ) then the electrons gets just enough quantum energy to come out of the metal. It means that the Kinetic Energy of such an electron will be zero. So we can write that -
$$
\mathrm{hv}_0-\Phi=0 \text { or } \mathrm{hv}_0=\Phi
$$
This is the relation between the threshold frequency and the work function. We can also change this equation in terms of the threshold wavelength.
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