Question

Find the de Broglie wavelength of electrons in hydrogen’s n= 1 state,n= 4 state, andn= 10 state

Answer 1

Answer:

Wavelength is 0.33 nm for n=1, 1.32 nm for n=4, and 3.3 nm for n=10.

Explanation:

The velocity of the electron in the nth level of hydrogen atom is,

$v_{n}=\frac{2.2\times 10^{6} m/s }{n}$

Now the de Broglie wavelength can be calculated as,

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

For n=1 the wavelength is,

$\lambda=\frac{6.626\times 10^{-34}Js }{9.1\times 10^{-31}kg(2.2\times 10^{6} m/s) }\\\lambda=0.33\times 10^{-9}m\\ \lambda=0.33nm$

Therefore wavelength for n=1 orbit is 0.33 nm.

For n=4 the wavelength is,

$\lambda=\frac{6.626\times 10^{-34}Js }{9.1\times 10^{-31}kg(\frac{2.2\times 10^{6} m/s}{4} ) }\\\lambda=1.32\times 10^{-9}m\\ \lambda=1.32nm$

Therefore wavelength for n=4 orbit is 1.32 nm.

For n=4 the wavelength is,

$\lambda=\frac{6.626\times 10^{-34}Js }{9.1\times 10^{-31}kg(\frac{2.2\times 10^{6} m/s}{4} ) }\\\lambda=3.3\times 10^{-9}m\\ \lambda=3.3nm$

Therefore wavelength for n=10 orbit is 3.3 nm.

Answer 2

Answer:

The de Broglie wavelength for n = 1

$\lambda=0.33\ nm$

The de Broglie wavelength for n = 4

$\lambda=1.33\ nm$

The de Broglie wavelength for n = 10

$\lambda=3.33\ nm$

Explanation:

Given that,

State of hydrogen’s atom are

n = 1,4 and 10

We need to calculate the energy for $n^{th}$ state

Using formula of energy

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

For , n = 1

$E=\dfrac{-13.6\times1.6\times10^{-19}}{1^2}}$

$E_{1}=-2.176\times10^{-18}\ J$

For, n = 4

$E=\dfrac{-13.6\times1.6\times10^{-19}}{4^2}}$

$E_{4}=-1.36\times10^{-19}\ J$

For, n = 10

$E=\dfrac{-13.6\times1.6\times10^{-19}}{10^2}}$

$E_{10}=-2.176\times10^{-20}\ J$

We need to calculate the de Broglie wavelength of electrons in hydrogen’s atom

Using formula of wavelength

$\lambda=\dfrac{h}{\sqrt{2mE}}$

For,n =1

$\lambda=\dfrac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times(2.176\times10^{-18})}}$

$\lambda=3.33\times10^{-10}\ m$

$\lambda=0.33\ nm$

For, n=4

$\lambda=\dfrac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times(1.36\times10^{-19})}}$

$\lambda=1.33\times10^{-9}\ m$

$\lambda=1.33\ nm$

For, n=10

$\lambda=\dfrac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times(2.176\times10^{-20})}}$

$\lambda=3.33\times10^{-9}\ m$

$\lambda=3.33\ nm$

Hence, The de Broglie wavelength for n = 1

$\lambda=0.33\ nm$

The de Broglie wavelength for n = 4

$\lambda=1.33\ nm$

The de Broglie wavelength for n = 10

$\lambda=3.33\ nm$