rohanprabhu
- 410
- 2
For the photoelectric effec to take place for a certain metal, there is a minimum frequency that the incident radiation must have, also called the threshold frequency. Let us call it f_o.
The electron is emmitted only when a minimum amount of energy is provided to it so that it can repulse the intermolecular forces. This energy, called the 'Work function' is \phi = f_o [where 'h' is the Planck's constant].
Now, if the minimum energy is not provided, the electron is not emitted. No matter how high the intensity of the incident radiation is, the electron won't be emitted unless the radiation has a frequency f_o.
But, if it is energy dependent, won't increasing the Intensity increase the energy too? Because,
I = \frac{dP}{dA}
Therefore, if a radiation has a higher intensity, won't it be having a higher value of Power? Which means that the energy supplied per unit time should also be higher. So, for a higher value of Intensity, the energy is more... So why doesn't it show the photoelectric effect unless and until the metal is irradiated with a radiation having a frequency of f_o
The electron is emmitted only when a minimum amount of energy is provided to it so that it can repulse the intermolecular forces. This energy, called the 'Work function' is \phi = f_o [where 'h' is the Planck's constant].
Now, if the minimum energy is not provided, the electron is not emitted. No matter how high the intensity of the incident radiation is, the electron won't be emitted unless the radiation has a frequency f_o.
But, if it is energy dependent, won't increasing the Intensity increase the energy too? Because,
I = \frac{dP}{dA}
Therefore, if a radiation has a higher intensity, won't it be having a higher value of Power? Which means that the energy supplied per unit time should also be higher. So, for a higher value of Intensity, the energy is more... So why doesn't it show the photoelectric effect unless and until the metal is irradiated with a radiation having a frequency of f_o