Question

For a lead ${ } _ { 82 } ^ { 208 } \mathrm {~Pb}$μ−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_λ = ∆E = E_i – E_f

$\alpha$= mZ² E₀ $\left[ \frac { 1 } { 1 ^ { 2 } } - \frac { 1 } { 2 ^ { 2 } } \right]$

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

= 19 MeV × 3/4 = 14 MeV