
The work function of sodium is 2.3 eV. The threshold wavelength of sodium will be
A. \[2900\mathop A\limits^ \circ \]
B. \[2500\mathop A\limits^ \circ \]
C. \[5380\mathop A\limits^ \circ \]
D. \[2000\mathop A\limits^ \circ \]
Answer
163.8k+ views
Hint: In order to answer this question, we have to apply the concept of photoelectric effect. According to this effect, if the frequency of incident radiation is greater than the threshold frequency then the photoemission of the electron will occur.
Formula used:
\[E = h\nu \]
where h is the Plank’s constant and E is the energy of the photon with frequency equal to \[\nu \]
\[c = \nu \lambda \]
where c is the speed of light, \[\nu \]is the frequency of the photon and \[\lambda \]is the wavelength of the light wave.
Complete step by step solution:
The photon is the qualitative unit of energy of the light wave. It is proportional to the frequency of the light wave. The work function of the sodium is given as 2.3 eV. So, the work function of the sodium in S.I. unit is,
\[\phi = 2.3 \times 1.6 \times {10^{ - 19}}J\]
\[\Rightarrow \phi = 3.68 \times {10^{ - 19}}J\]
The frequency corresponding to the energy of the work function is called the threshold frequency.
\[\phi = h{\nu _0}\]
\[\Rightarrow {\nu _0} = \dfrac{\phi }{h}\]
Using the relation between the speed of light, frequency and the wavelength, we get
\[\dfrac{c}{{{\lambda _0}}} = \dfrac{\phi }{h}\]
\[\Rightarrow {\lambda _0} = \dfrac{{hc}}{\phi }\]
Now putting the values of the above quantities we get,
\[{\lambda _0} = \dfrac{{\left( {6.626 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{3.68 \times {{10}^{ - 19}}}}m\]
\[\Rightarrow {\lambda _0} = 5.402 \times {10^{ - 7}}m\]
\[\Rightarrow {\lambda _0} = 5.402 \times {10^3} \times {10^{ - 10}}m\]
\[\Rightarrow {\lambda _0} = 5.402 \times {10^3}\mathop A\limits^ \circ \]
\[\therefore {\lambda _0} \simeq 5400\mathop A\limits^ \circ \]
Hence, the threshold wavelength of the sodium is \[5400\mathop A\limits^ \circ\].
Therefore, the correct option is C, which is nearest to the available options
Note: We must be very careful about the difference between threshold frequency and frequency of incident radiation. The threshold frequency is characteristic of the metal and independent of the intensity of incident radiation while the incident frequency is dependent on the energy of the incident radiation.
Formula used:
\[E = h\nu \]
where h is the Plank’s constant and E is the energy of the photon with frequency equal to \[\nu \]
\[c = \nu \lambda \]
where c is the speed of light, \[\nu \]is the frequency of the photon and \[\lambda \]is the wavelength of the light wave.
Complete step by step solution:
The photon is the qualitative unit of energy of the light wave. It is proportional to the frequency of the light wave. The work function of the sodium is given as 2.3 eV. So, the work function of the sodium in S.I. unit is,
\[\phi = 2.3 \times 1.6 \times {10^{ - 19}}J\]
\[\Rightarrow \phi = 3.68 \times {10^{ - 19}}J\]
The frequency corresponding to the energy of the work function is called the threshold frequency.
\[\phi = h{\nu _0}\]
\[\Rightarrow {\nu _0} = \dfrac{\phi }{h}\]
Using the relation between the speed of light, frequency and the wavelength, we get
\[\dfrac{c}{{{\lambda _0}}} = \dfrac{\phi }{h}\]
\[\Rightarrow {\lambda _0} = \dfrac{{hc}}{\phi }\]
Now putting the values of the above quantities we get,
\[{\lambda _0} = \dfrac{{\left( {6.626 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{3.68 \times {{10}^{ - 19}}}}m\]
\[\Rightarrow {\lambda _0} = 5.402 \times {10^{ - 7}}m\]
\[\Rightarrow {\lambda _0} = 5.402 \times {10^3} \times {10^{ - 10}}m\]
\[\Rightarrow {\lambda _0} = 5.402 \times {10^3}\mathop A\limits^ \circ \]
\[\therefore {\lambda _0} \simeq 5400\mathop A\limits^ \circ \]
Hence, the threshold wavelength of the sodium is \[5400\mathop A\limits^ \circ\].
Therefore, the correct option is C, which is nearest to the available options
Note: We must be very careful about the difference between threshold frequency and frequency of incident radiation. The threshold frequency is characteristic of the metal and independent of the intensity of incident radiation while the incident frequency is dependent on the energy of the incident radiation.
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