The emission of electrons in a photodiode is possible when the energy of the incident radiation is greater than the forbidden energy gap (Eg) of the material used. The energy (E) of the radiation can be calculated using the formula E = hν, where h is Planck's constant (approximately 4.14 x 10^-15 eV.s) and ν is the frequency of the radiation.

Given that Eg = 3.7 eV and ν = 1 MHz, we can calculate the energy of the radiation:

E = hν = (4.14 x 10^-15 eV.s) x (1 x 10^6 s^-1) = 4.14 x 10^-9 eV

This is much less than the forbidden energy gap (Eg = 3.7 eV), so the emission of electrons is not possible with a radiation frequency of 1 MHz.

Therefore, for the emission of electrons to be possible, the frequency of the radiation must be greater than 1 MHz. So, the correct answer is:

c. ν 1 MHz

Question

The emission of electrons in a photodiode is possible when the energy of the incident radiation is greater than the forbidden energy gap (Eg) of the material used. The energy (E) of the radiation can be calculated using the formula E = hν, where h is Planck's constant (approximately 4.14 x 10^-15 eV.s) and ν is the frequency of the radiation.

Given that Eg = 3.7 eV and ν = 1 MHz, we can calculate the energy of the radiation:

E = hν = (4.14 x 10^-15 eV.s) x (1 x 10^6 s^-1) = 4.14 x 10^-9 eV

This is much less than the forbidden energy gap (Eg = 3.7 eV), so the emission of electrons is not possible with a radiation frequency of 1 MHz.

Therefore, for the emission of electrons to be possible, the frequency of the radiation must be greater than 1 MHz. So, the correct answer is:

c. ν > 1 MHz

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The emission of electrons in a photodiode is possible when the energy of the incident radiation is greater than the forbidden energy gap (Eg) of the material used. The energy (E) of the radiation can be calculated using the formula E = hν, where h is Planck's constant (approximately 4.14 x 10^-15 eV.s) and ν is the frequency of the radiation.

Given that Eg = 3.7 eV and ν = 1 MHz, we can calculate the energy of the radiation:

E = hν = (4.14 x 10^-15 eV.s) x (1 x 10^6 s^-1) = 4.14 x 10^-9 eV

This is much less than the forbidden energy gap (Eg = 3.7 eV), so the emission of electrons is not possible with a radiation frequency of 1 MHz.

Therefore, for the emission of electrons to be possible, the frequency of the radiation must be greater than 1 MHz. So, the correct answer is:

c. ν > 1 MHz

For a photodiode, the forbidden energy gap (Eg) for the material used is 3.7 eV and frequency of radiation (ν) is1 MHz, then the emission of electrons is possible when incident radiation have,Select one:a. ν = 1 MHzb. ν < 1 MHzc. ν > 1 MHzd. ν = 500 kHz

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