Question

Find out ratio of following for photon $(v_{max}{)}_{Lyman}:(v_{max}{)}_{Brakett}$.

• A. $1:16$
• B. $16:1$
• C. $4:1$
• D. $1:4$

Answer 1

Answer: (C)

$4:1$

Maximum Kinetic Energy is given by the expression:

$\frac{1}{2}\times m{v}_{max}^2=e{V}_o=\frac{hc}{\Lambda }$

$\frac{1}{\Lambda }=\frac{R{Z}^2}{n_1^2}$

$\frac{1}{2}\times m{v}_{max}^2=\frac{hc}{\Lambda }=\frac{R{Z}^{2}hc}{n_1^2}$

$v_{max}^2=\frac{2R{Z}^{2}hc}{m{n}_1^2}$

$v_{max}=\sqrt{\frac{2R{Z}^{2}hc}{m{n}_1^2}}$

$v_{max}\alpha \frac{1}{n_1}$

For Lyman series: n1 = 1 and For Brakett series: n1 = 4

$(v_{max}{)}_{Lyman}:(v_{max}{)}_{Brakett}=4:1$