
For photoelectric emission, tungsten requires light of \[2300\mathop A\limits^ \circ \]. If light of \[1800\mathop A\limits^ \circ \] wavelength is incident, then emission:
A. Takes place
B. Doesn’t take place
C. May or may not take place
D. Depends on frequency
Answer
164.1k+ views
Hint: The threshold frequency of the metal is the frequency corresponding to the minimum energy that is required to eject the electron from the surface of metal.
Formula used:
\[K = h\nu - {\phi _0}\]
Here K is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, \[\nu \] is the frequency of the photon and \[\phi \] is the work function of the metal.
\[c = \nu \lambda \]
Here c is the speed of light, \[\nu \] is the frequency and \[\lambda \] is the wavelength of the electromagnetic wave.
Complete step by step solution:
For the minimum condition, the kinetic energy of the ejected electron is zero.
Using the energy formula,
\[{\phi _0} = h{\nu _0}\]
Using the relation between the frequency and the wavelength,
\[{\phi _0} = \dfrac{{hc}}{{{\lambda _0}}}\]
Putting in the formula of the kinetic energy, we get
\[K = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}}\]
\[K = hc\left( {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}}} \right)\]
For the emission of the electron, the kinetic energy of the ejected electron must be non-zero. If the energy of the photon exceeds the minimum energy needed to eject the electron then the rest of the energy is transferred as kinetic energy of the ejected electrons.
So,
\[K \ge 0\]
\[hc\left( {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}}} \right) \ge 0\]
\[\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}} \ge 0\]
\[\dfrac{1}{\lambda } \ge \dfrac{1}{{{\lambda _0}}}\]
\[\lambda \le {\lambda _0}\]
As per the given data \[{\lambda _0} = 2300\mathop A\limits^ \circ \] and \[\lambda = 1800\mathop A\limits^ \circ \]
Hence, the emission takes place.
Therefore, the correct option is A.
Note: It is important to remember that the maximum kinetic energy of an ejected electron is directly proportional to the frequency of incident radiation and independent to the intensity of incident radiation.
Formula used:
\[K = h\nu - {\phi _0}\]
Here K is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, \[\nu \] is the frequency of the photon and \[\phi \] is the work function of the metal.
\[c = \nu \lambda \]
Here c is the speed of light, \[\nu \] is the frequency and \[\lambda \] is the wavelength of the electromagnetic wave.
Complete step by step solution:
For the minimum condition, the kinetic energy of the ejected electron is zero.
Using the energy formula,
\[{\phi _0} = h{\nu _0}\]
Using the relation between the frequency and the wavelength,
\[{\phi _0} = \dfrac{{hc}}{{{\lambda _0}}}\]
Putting in the formula of the kinetic energy, we get
\[K = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}}\]
\[K = hc\left( {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}}} \right)\]
For the emission of the electron, the kinetic energy of the ejected electron must be non-zero. If the energy of the photon exceeds the minimum energy needed to eject the electron then the rest of the energy is transferred as kinetic energy of the ejected electrons.
So,
\[K \ge 0\]
\[hc\left( {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}}} \right) \ge 0\]
\[\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}} \ge 0\]
\[\dfrac{1}{\lambda } \ge \dfrac{1}{{{\lambda _0}}}\]
\[\lambda \le {\lambda _0}\]
As per the given data \[{\lambda _0} = 2300\mathop A\limits^ \circ \] and \[\lambda = 1800\mathop A\limits^ \circ \]
Hence, the emission takes place.
Therefore, the correct option is A.
Note: It is important to remember that the maximum kinetic energy of an ejected electron is directly proportional to the frequency of incident radiation and independent to the intensity of incident radiation.
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