Question

Energy of electron in orbit of H-atom is -1.51 eV
Wavelength produced by the electron in same orbit
if 1st orbit of H have radius x
(1) 2nex
(2) 3rx
(3) 6rx
​

Answer 1

Answer:

Wavelength produced by the electron in same orbit

$\lambda = 6\pi x$

Explanation:

As we know that energy of electron for hydrogen atom in its nth orbit is given as

$E = - 13.6 \frac{1}{n^2} eV$

now we know that

$E = -1.51 eV$

so we will have

$-1.51 = -13.6 \frac{1}{n^2}$

so we have

$n = 3$

now we know that radius of orbit is given as

$r = 0.529 n^2 A^o$

so we will have n = 1 for 1st orbit

$x = 0.529 A^o$

so we can say for given orbit we have

$r = x(3^2)$

$r = 9x$

now wavelength of the electron in this orbit is given as

$\lambda = \frac{h}{mv}$

also we know that

$mvr = \frac{nh}{2\pi}$

so we have

$\lambda = \frac{h}{mv} = \frac{2\pi r}{n}$

here we know

r = 9x

n = 3

$\lambda = \frac{2\pi 9x}{3}$

$\lambda = 6\pi x$

#Learn

Topic : De Broglie Wavelength

https://brainly.in/question/1489844