Question

The wavelength of the de-Broglie wave of an electron revolving in the sixth orbit of the hydrogen is: ($r_0$ is the Bohr's radius=$0.529\dot{A}$)

• A. $937.62\dot{A}$
• B. $119.6\dot{A}$
• C. $235.67\dot{A}$
• D. none of these

Answer 1

Answer: (D)

none of these

$r=0.0529\times \frac{n^2}{z}$

$r=0.529\times \frac{36}{1}\mathring{A}=36\times 0.529\mathring{A}=19.044\mathring{A}$

Now,

$2nr=6\times \lambda$

$\Rightarrow \lambda =\frac{\pi }{3}\times 19.044$

$=19.94\mathring{A}$