Question

For a Lead ${}_{82}^{208}Pb$m-mesonic atom the energy of the photon given off in the first lyman transition i.e. (n = 2 to n = 1)

• A. 11 MeV
• B. 14 MeV
• C. 18 MeV
• D. 20 MeV

Answer 1

Answer: (B)

14 MeV

Answer 2

E_l = DE = E_i – E_f

= mz²E₀ $\left[ \frac { 1 } { 1 ^ { 2 } } - \frac { 1 } { 2 ^ { 2 } } \right]$

= (207) (82)² × 13.6 $\left[ \frac { 1 } { 1 ^ { 2 } } - \frac { 1 } { 2 ^ { 2 } } \right]$

= 19MeV × $\frac { 3 } { 4 }$ = 14 MeV