Answer
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Hint: We know that the threshold wavelength is defined as the maximum wavelength that the incident radiation must have so that the photoelectric effect takes place. We should also know that the frequency of the incident light is less than the threshold frequency. No photoelectrons will be emitted from the metal surface. Based on this concept we have to answer this question.
Complete step by step answer:
At first let us mention the values of the questions that are mentioned to us:
Threshold wavelength =$5,200{{A}^{0}}$which can also be written as 520nm
We know that the formula of energy is given as: $\dfrac{he}{\lambda }$
In the above expression;
h is the Planck constant; e is the charge of the electron and $\lambda$is the wavelength.
So, now we have evaluated to get the value as:
$=\dfrac{1240}{520}eV\ \left( he=1240eV-nm \right)$
$=2.38eV$
So, the threshold energy is $=2.38eV$
So, for ejection of electrons energy of incident rays should be more than $2.38eV$.
Hence, the correct answer is Option C.
Note: We know that electromagnetic radiation, especially visible radiation of only one frequency or wavelength. Completely monochromatic radiation cannot be produced, but lasers produce radiation within a very narrow frequency band.
It should also be known to us that the term monochromatic originally means having only one single colour. Different optical wavelengths of the visible light are associated with the different perceived colours. However, light colours are rarely a criterion for monochromatically in the practice, and the non-monochromatic light can also have specific colours.
Complete step by step answer:
At first let us mention the values of the questions that are mentioned to us:
Threshold wavelength =$5,200{{A}^{0}}$which can also be written as 520nm
We know that the formula of energy is given as: $\dfrac{he}{\lambda }$
In the above expression;
h is the Planck constant; e is the charge of the electron and $\lambda$is the wavelength.
So, now we have evaluated to get the value as:
$=\dfrac{1240}{520}eV\ \left( he=1240eV-nm \right)$
$=2.38eV$
So, the threshold energy is $=2.38eV$
So, for ejection of electrons energy of incident rays should be more than $2.38eV$.
Hence, the correct answer is Option C.
Note: We know that electromagnetic radiation, especially visible radiation of only one frequency or wavelength. Completely monochromatic radiation cannot be produced, but lasers produce radiation within a very narrow frequency band.
It should also be known to us that the term monochromatic originally means having only one single colour. Different optical wavelengths of the visible light are associated with the different perceived colours. However, light colours are rarely a criterion for monochromatically in the practice, and the non-monochromatic light can also have specific colours.
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