Electron in a hydrogen-like atom `(Z = 3)` make transition from the forth excited state to the third excited state and from the third excited state to the second excited state. The resulting radiations are incident potential for photoelectrons ejested by shorter wavelength is `3.95 e V`.
Calculate the work function of the metal and stopping potiential for the photoelectrons ejected by the longer wavelength.

Explanation

Expression of wavelength for hydrogen-like atoms can be written as follows :
`1/lambda=RZ^2[1/n_1^2-1/n_2^2]`
First wavelength :
`1/lambda_1=(1.094xx10^7)xx(3)^2xx[1/4^2-1/5^2]=0.2215xx10^7 m^(-1)`
Second wavelength :
`1/lambda_2=(1.094xx10^7)xx(3)^2xx[1/3^2-1/4^2]=0.4786xx10^7 m^(-1)`
Here second wavelength is shorter for which stopping potential is given to us.
According to photoelectric equation we can write the following :
`(hc)/lambda=W+eV_0`
Here W is work function of the material and `V_0` is stopping potential .
As per given information we can write the following :
`(hc)/lambda=W+exx3.95`
`rArr W=(hc)/lambda-3.95e`
`rArr W=6.63xx10^(-34)xx3xx10^8xx0.4786xx10^7-3.95xx1.6xx10^(-19)`
`rArr W=9.519xx10^(-19)-6.32xx10^(-19)`
`rArr W=3.2xx10^(-19)J =(3.2xx10^(-19))/(1.6xx10^(-19))`eV=2eV
Hence, we have calculated work function of the metal. Now we proceed to calculate stopping potential corresponding to the other wavelength in the radiation.
`(hc)/lambda_2=W+eV_0`
`6.63xx10^(-34)xx3xx10^8xx0.2215xx10^7`
`=3.2xx10^(-19)+1.6xx10^(-19)xxV_0`
`4.4xx10^(-19)=3.2xx10^(-19)+1.6xx10^(-19)xxV_0`
`V_0=1.2/1.6`=0.75 V
Hence stopping potential for longer wavelength will be 0.75 V