Question

Find the wavelength of the most energetic photons in this series.

• A. $1800\mathring{A}$
• B. $1500\mathring{A}$
• C. $1300\mathring{A}$
• D. $1650\mathring{A}$

Answer 1

Answer: (B)

$1500\mathring{A}$

Given, wavelength is, $\lambda =\frac{1500p^2}{p^2-1}wherep=2,3,\mathrm{4....}$
$\Rightarrow \lambda =1500(1+\frac{1}{p^2-1})$
that implies wavelength decreases with increase in $p$.

As, Energy $\propto \frac{1}{\lambda }$
Most energetic photons have least wavelength.
Therefore, wavelength of the most energetic photons is:
${limit}_{p⟶\infty }\lambda ={limit}_{p⟶\infty }1500(1+\frac{1}{p^2-1})$
And, ${limit}_{p⟶\infty }(1+\frac{1}{p^2-1})=1$
Therefore, required wavelength is $1500\mathring{A}$.