Before solving this question first we have to understand work function.
The work function of a metal is amount of minimum energy required to emit an electron from the surface barrier of metal . Whenever the metal will be exposed to radiation a part of its energy will be utilized to emit an electron while rest will provide kinetic energy to the electron.
Let f is the frequency of incident radiation and f' is the frequency corresponding to work function. Let v is the velocity of the ejected electron.
we know that velocity of an electromagnetic wave is the product of frequency and wavelength. Hence frequency f is given as-
where c is velocity of light and is the wavelength of the wave.
As per the question incident wavelength =313 nm
[as 1 nm =10^-9 m]
The wavelength corresponding to work function is 351 nm i.e
we know that hf=hf'+K.E [ h is the planck's constant whose value is 6.63×10^-34 J-s]
⇒K.E =hf-hf'
[ans]
I think it is 5% hope this helped
Answer:
you couldn't do this on your own or search it up on google
I just figured this out now.
First you would use the formula
Ephoton= hc/λ and substitute in the value's of plank's constant, the speed of light in a vaccum and the wavelength which will give you the energy in joules. Then you go to the reference table and solve for the energy used between the different levels for Mercury making sure to convert electron volts to jules. In the end the correct answer should be energy level D.
Answer:
there is an mobile an a latter