Distinguish between nuclear fission and fusion. Show how in both these processes energy is …… Calculate the energy release in MeV in the deuteriumtritium fusion reaction:
`._(1)^(2)H+._(1)^(3)Hrarr._(1)^(3)He+n`
Using the data.
`{:(m(._(1)^(2)H)=2.014102 u,m(._(1)^(3)H)=3.016049 u),(m(._(1)^(3)He)=4.002603 u,m_(n)=1.008665 u),(lu=931.5 Me V//c^(2),):}`

Explanation

Elnstein’s photoelectric equation:
Where v = incident frequency, `v_(0 =` threshold frequency, `V_(0) =` stopping potential
(i) Incident energy of photon is used in to (a) to liberate electron from the metal surface (b) rest of the energy appears as maximum energy electron.
(ii) Only one electron can absorb energy of one photon. Hence increasing intensity increases the numbers of electrons hence current.
(iii) If incident energy is less than work function, no emission of electron will take place.
(iv) Increasing v (incident frequency) will increase maximum kinetic energy of electrons but number of electrons emitted will remain same.
For wavelength
`(hc)/(lambda_(1))=phi_(0)+K`
`= phi_(0)+ ._(e )v_(0) " "` .... (i) where `K = ._(e )v_(0)`
Form wavelength `lambda_(2)`
`(hc)/(lambda_(1))=phi_(0)+2_(e )v_(0)` ....(ii) (because KE is doubled)
Form equartion (i) and (ii), we get
`(hc)/(lambda_(2))=f_(0)+2((hc)/(lambda_(1))-phi_(0))`
`= phi_(0)+(2hc)/(lambda_(1))-2phi_(0)`
`rArr " " lambda_(0)=(2hc)/(lambda_(1))-(hc)/(lambda_(2))`
For thershold wavelength `lambda_(0)`, kinetic energy, K = 0
and work function `lambda_(0)=(hc)/(lambda_(0))`
`therefore (hc)/(lambda_(0))=(2hc)/(lambda_(1))-(hc)/(lambda_(0))`
`rArr " " (1)/(lambda_(0))=(2)/(lambda_(1))-(1)/(lambda_(2))`
`rArr " " lambda_(0)=(lambda_(1)lambda_(2))/(2lambda_(2)-lambda_(1))`
Work function, `lambda_(0)=(hc(2lambda_(2)-lambda_(1)))/(lambda_(1)lambda_(2))`