A photon of wavelength 3310Å falls on a photocathode and an electron of energy `3xx10^(-19)J` is ejected. If the wavelength of the incident photon is changed to 5000Å, the energy of the ejected electron is `9.72xx10^(-20)J`. Calculate the value of Planck's constant and threshold wavelength of the photon.

Explanation

When a photon of wavelength `lamda` is incident on a photocathode , for which threshold wavelength is `lamda_0`, the energy of the ejected electron is
`E=(hc)/lamda-(hc)/lamda_0`
Given, `E_1=3xx10^(-19)J`
and `lamda_1=3310Å = 3310 xx10^(10)m`
`:. 3xx10 ^(-19)=(hc)//(3310xx10^(-10) )-(hc)/lamda_0" "...(i) `
`E_2=9.72xx10^(-20)J`
`lamda_2=500Å=500xx10^(-10)m`
`:. 9.72xx10^(-20)=(hc)/(5000xx10^(-10))-(hc)/lamda_0`
`implies 0.972xx10^(-19)=(hc)/(5000xx10^(-10))-(hc)/lamda_0" "...(ii)`
On subtracting (ii) from (i) , we get
`(3-0.972)xx10^(-19)=(hc)/(3310xx10^(-10))-(hc)/(5000xx10^(-10))`
`=(hc)/(10^(-10))(1/3310-1/5000)`
`implies2.028xx10^(-19)=(hc)/(10^(-10))=(1690/(3310xx5000))`
`impliesh=(2.028xx10^(-19)xx10^(-10)xx3310xx5000)/(3xx10^8xx1690)`
`=6.62xx10^(-34)JS`
Using relation,
`E=(hc)/lamda-(hc)/lamda_0`
Or `(hc)/lamda_0=(hc)/lamda-E`
where `lamda_0` is the threshold frequency.
Then
`(hc)/lamda_0=(hc)/lamda-E=(6.62xx10^(-34)xx3xx10^8)/(3310xx10^(-19))-3xx10^(-19)`
`=3xx10^(-19)J`
`implieslamda_0=(hc)/(9xx10^(-19))=(6.62xx10^(-34)xx3xx10^8)/(3xx10^(-19))`
`=6.63xx10^(-7)m`