Question

what is the minimum frequency of photon that can eject photoeletron from Ba if it's work function is 4.3*10^-19 J​

Answer 1

Answer:

Explanation:

The frequency of radiation corresponding to the energy 3.97×10  

−19

J is given by -

υ=E/h where E = energy , h= planck's constant

Also we know that,  

1J=10  

7

erg

so, 3.97×10  

−19

J=3.97×10  

−12

erg

Therefore, υ=  

6.626×10  

−27

erg−sec

3.97×10  

−12

erg

​  

 

υ=5.99\times 10^{14} Hz

(b) Energy corresponding to the wavelength 450 nm :

We know E=hc/λ

where E = energy , h= planck's constant , λ = wavelength

E=  

450×10  

−9

m

6.626×10  

−27

erg−sec×3×10  

8

m

​  

 

E=4.41×10  

−12

erg

Since this energy is greater than the given threshold energy 3.97×10  

−12

erg, thus the blue light of the given wavelength will be able to eject the electrons.