
Ultraviolet radiation of 6.2eV falls on an aluminium surface (work function 4.2eV). Then find the kinetic energy (in joule) of the fastest electron emitted.
A. \[3.2 \times {10^{ - 21}}J\]
B. \[1.6 \times {10^{ - 17}}J\]
C. \[3.2 \times {10^{ - 19}}J\]
D. \[3.2 \times {10^{ - 15}}J\]
Answer
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Hint: When light is incident on a metal surface, electrons are emitted known as photoelectrons. This phenomenon is called the photoelectric effect. Work function is the minimum energy required to remove one electron from a metal surface. Work function can also be said as the characteristics of the surface and not the complete metal. As we know that the energy of the incident light or photons should be more than the work function of the metal for the photoemission to take place.
Formula Used:
To find the kinetic energy of the emitted electron we have,
\[K.E = E - W\]
Where, W is the work function and E is the energy of a photon.
Complete step by step solution:
As we know that the kinetic energy of the emitted electron is,
\[K.E = E - W\]
Here, the energy of radiation is 6.2eV and the work function is 4.2eV. Since both the values are in eV, we need to convert it into joules by multiplying it by \[1.6 \times {10^{ - 19}}J\]. Then,
\[E = 6.2 \times 1.6 \times {10^{ - 19}}J\]
\[\Rightarrow E = 9.92 \times {10^{ - 19}}J\]and
\[\Rightarrow W = 4.2 \times 1.6 \times {10^{ - 19}}J\]
\[\Rightarrow W = 6.72 \times {10^{ - 19}}J\]
Now, substitute these value sin above equation, we get,
\[K.E = 9.92 \times {10^{ - 19}} - 6.72 \times {10^{ - 19}}\]
\[\therefore K.E = 3.2 \times {10^{ - 19}}J\]
Therefore, the kinetic energy of the fastest electron emitted is \[3.2 \times {10^{ - 19}}J\].
Hence, Option C is the correct answer.
Note:Remember that, if the energy is given in eV, then we need to convert the given value of energy from eV to joules by multiplying the charge of an electron while solving this problem. Moreover, Metals contain many free electrons that are loosely bound to the atom. The typical value of work function varies from 2ev to 6eV.
Formula Used:
To find the kinetic energy of the emitted electron we have,
\[K.E = E - W\]
Where, W is the work function and E is the energy of a photon.
Complete step by step solution:
As we know that the kinetic energy of the emitted electron is,
\[K.E = E - W\]
Here, the energy of radiation is 6.2eV and the work function is 4.2eV. Since both the values are in eV, we need to convert it into joules by multiplying it by \[1.6 \times {10^{ - 19}}J\]. Then,
\[E = 6.2 \times 1.6 \times {10^{ - 19}}J\]
\[\Rightarrow E = 9.92 \times {10^{ - 19}}J\]and
\[\Rightarrow W = 4.2 \times 1.6 \times {10^{ - 19}}J\]
\[\Rightarrow W = 6.72 \times {10^{ - 19}}J\]
Now, substitute these value sin above equation, we get,
\[K.E = 9.92 \times {10^{ - 19}} - 6.72 \times {10^{ - 19}}\]
\[\therefore K.E = 3.2 \times {10^{ - 19}}J\]
Therefore, the kinetic energy of the fastest electron emitted is \[3.2 \times {10^{ - 19}}J\].
Hence, Option C is the correct answer.
Note:Remember that, if the energy is given in eV, then we need to convert the given value of energy from eV to joules by multiplying the charge of an electron while solving this problem. Moreover, Metals contain many free electrons that are loosely bound to the atom. The typical value of work function varies from 2ev to 6eV.
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