Question

Find the principle quantum number for the electron having the de broglie wavelength in an orbit of hydrogen atom is 10^-9 m.
(Final ans is 3; please explain with steps)

Answer 1

Given lambda=10^-9
h/MV = 10^-9
we know MVR=nh/2π
Thus, replacing value of MV in de brogle equation we get,
2πr/n = 10^-9
Now for hydrogen like atoms, r=0.053*10^-10*n^2
putting the value of r in above equation and then solving, you'll get 3 as answer. I've done it.

Answer 2

The principle quantum number for the electron is 3.

Given:
De- Broglie wavelength $=10^{-9} m$

To find:

Principal quantum number

Solution:

Step 1

The de- Broglie wavelength for an electron revolving in an orbit of hydrogen atom is given by

λ  $=\frac{h}{P}$

We know,

λ $=10^{-9}m$

$h=6.634$ × $10^{-34} m$

Substituting the given values in the equation, we get

$10^{-9} =\frac{6.634*10^{-34} }{P}$

$P=6.634$ × $10^{-25} kgm/s$

Step 2

The momentum $(P)$ of any body of mass $(m)$ and velocity $(v)$ in motion is given by,

$P=mv$

We know, the velocity of any particle revolving inside the atom is given by

$v=2.1$ × $10^{6} \frac{Z}{n}$

Where, $Z$ is the atomic number and $n$ is the principal quantum number of the orbit of the atom in which the particle such as electron is revolving.

Hence, we get the equation of momentum for the electron as

$P=m$ × ( $2.1$ × $10^{6} .\frac{Z}{n}$ )

Step 3

We know,

Mass of electron $=9.1$ × $10^{-31}kg$

Atomic number $Z$ of hydrogen atom $=1$

Momentum $P=6.6*10^{-25}$ $kg.m/s$

Substituting the given values in the equation, we get

$6.6*10^{-25} = \frac{(9.1 * 10^{-31} )*(2.1*10^{6}).(1) }{n}$

$n=\frac{19.11*10^{-25} }{6.6*10^{-25} }$

$n=\frac{19.11}{6.6}$

$n$ ≈ $3$

Final answer:

Hence, the principle quantum number for the electron is 3.