Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
For a certain metal, when monochromatic light of wavelength $\lambda$ is incident, the stopping potential for photoelectrons is $3V_o$. When the same metal is illuminated by light of wavelength $2\lambda$, then the stopping potential becomes $V_o$. The threshold wavelength for photoelectric emission for the given metal is $\alpha\lambda$. The value of $\alpha$ is _______.
(1) 1
(2) 4
(3) 2
(4) 3
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Answer: Option (2)
Key idea: Photoelectric Effect - Einstein's Equation and Threshold Wavelength
Write Einstein's photoelectric equation for both cases: $\frac{hc}{\lambda} = \phi + 3eV_0$ and $\frac{hc}{2\lambda} = \phi + eV_0$
Subtract the second equation from the first: $\frac{hc}{\lambda} - \frac{hc}{2\lambda} = 2eV_0$
Simplify to get: $\frac{hc}{2\lambda} = 2eV_0$, therefore $hc = 4eV_0\lambda$
Substitute back into second equation: $2eV_0 = \phi + eV_0$, so work function $\phi = eV_0$
The graph shows variation of stopping potential $V_o$ with the frequency $v$ of the incident radiation for three photosensitive metals $X_1, X_2$ and $X_3$. Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation? [Figure: Graph showing three parallel lines labeled $X_1$, $X_2$, $X_3$ plotting $V_o$ versus $v$ (in $\times 10^{14}$ Hz), with frequency axis showing values 1.0, 1.5, 2]
(1) $X_1$
(2) $X_2$
(3) $X_3$
(4) All the metals will give out photo electrons with same kinetic energies.
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Answer: Option (1)
Key idea: Photoelectric Effect - Stopping Potential and Work Function Relationship
Write the photoelectric equation: $eV_0 = h\nu - \phi$, where $V_0$ is stopping potential and $\phi$ is work function
Note that maximum kinetic energy of photoelectrons is $KE_{max} = eV_0 = h\nu - \phi$
For the same incident frequency $\nu$, the metal with lowest work function $\phi$ gives highest kinetic energy
From the graph, parallel lines mean same slope $(h/e)$ but different work functions. Higher $V_0$ at given $\nu$ means lower $\phi$
At any given frequency, $X_1$ has the highest stopping potential $V_0$, meaning it has the lowest work function
Therefore, $X_1$ will give out electrons with the greatest kinetic energy for the same wavelength of incident radiation
The de Broglie wavelength associated with an electron accelerated through a potential difference V is $\lambda_e$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_p$. If their corresponding masses are $m_e$ and $m_p$, respectively, then the ratio of their de Broglie wavelengths $\left(\frac{\lambda_e}{\lambda_p}\right)$ is ______.
(1) $\sqrt{\frac{m_p}{m_e}}$
(2)
(3)
(4)
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Answer: Option (1)
Key idea: de Broglie wavelength and its dependence on mass for charged particles accelerated through same potential difference
When a charged particle is accelerated through potential difference $V$, it gains kinetic energy: $KE = eV$ (for electron and proton, charge magnitude is $e$)
The kinetic energy relates to momentum: $KE = \frac{p^2}{2m}$, giving $p = \sqrt{2meV}$
For electron: $\lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{2m_eeV}}$
For proton: $\lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2m_peV}}$
Taking the ratio: $\frac{\lambda_e}{\lambda_p} = \frac{\sqrt{2m_peV}}{\sqrt{2m_eeV}} = \sqrt{\frac{m_p}{m_e}}$
An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_o\hat{i}$ ($E_o$ = constant > 0), with an initial velocity $\vec{V} = v_o\hat{i}$ ($v_o$ = constant > 0). If $\lambda_o = \frac{h}{4mv_o}$, its de Broglie wavelength at time $t$ is ______. ($e$ = charge of electron)
Light source having wavelength $331$ nm is used to generate photo-electrons whose stopping potential is $0.2$ V. The work function of the used metal in the experiment is $a \times 10^{-19}$ J. The value of $a$ is ____. ($h = 6.62 \times 10^{-34}$ J s, $e = 1.6 \times 10^{-19}$ C and $c = 3 \times 10^8$ m/s)
An electron is travelling with a velocity $v$ in free space and when it enters a medium, its velocity is reduced by 20%. The de Broglie wavelength of electron in the medium is $\alpha\lambda_0$, where $\lambda_0$ is its de Broglie wavelength in free space. The value of $\alpha$ is ________.
(1) 1.20
(2) 1.0
(3) 1.25
(4)
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Answer: Option (3)
Key idea: de Broglie wavelength and its inverse relationship with velocity
Write the de Broglie wavelength formula: $\lambda = \frac{h}{mv}$
In free space: $\lambda_0 = \frac{h}{mv}$
Velocity in medium is reduced by 20%: $v_{medium} = 0.8v$
In medium: $\lambda_{medium} = \frac{h}{m(0.8v)} = \frac{h}{0.8mv}$
The de Broglie wavelength for an electron accelerated through the potential difference of $V_1$ volt is $\lambda_1$. When the potential difference is changed to $V_2$ volt, the associated de Broglie wavelength is increased by 50%. If $(V_1/V_2) = (9/a)$, then the value of $a$ is __________.
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Answer: 4
Key idea: de Broglie wavelength and its dependence on accelerating potential
For an electron accelerated through potential $V$, de Broglie wavelength $\lambda = \frac{h}{\sqrt{2meV}}$, so $\lambda \propto \frac{1}{\sqrt{V}}$
Given that wavelength increases by 50%, so $\lambda_2 = 1.5\lambda_1$
Using the relation $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}$, substitute $\lambda_2 = 1.5\lambda_1$
This gives $\frac{\lambda_1}{1.5\lambda_1} = \frac{2}{3} = \sqrt{\frac{V_2}{V_1}}$
Squaring both sides: $\frac{4}{9} = \frac{V_2}{V_1}$, so $\frac{V_1}{V_2} = \frac{9}{4}$
Comparing with $\frac{V_1}{V_2} = \frac{9}{a}$, we get $a = 4$
$K_1$ and $K_2$ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength $\lambda_1$ and $\lambda_2$, respectively. If $\lambda_1 = 2\lambda_2$ then the work function of material is given by:
