Question

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

Answer 1

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`

Step-by-step 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=(-13.6\ eV)/(n^2)}`

For , n = 1

`E=(-13.6*1.6*10^(-19))/(1^2)}`

`E_(1)=-2.176*10^(-18)\ J`

For, n = 4

`E=(-13.6*1.6*10^(-19))/(4^2)}`

`E_(4)=-1.36*10^(-19)\ J`

For, n = 10

`E=(-13.6*1.6*10^(-19))/(10^2)}`

`E_(10)=-2.176*10^(-20)\ J`

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

Using formula of wavelength

`\lambda=(h)/(√(2mE))`

For,n =1

`\lambda=\frac{6.63*10^(-34)}{\sqrt{2*9.1*10^(-31)*(2.176*10^(-18))}}`

`\lambda=3.33*10^(-10)\ m`

`\lambda=0.33\ nm`

For, n=4

`\lambda=\frac{6.63*10^(-34)}{\sqrt{2*9.1*10^(-31)*(1.36*10^(-19))}}`

`\lambda=1.33*10^(-9)\ m`

`\lambda=1.33\ nm`

For, n=10

`\lambda=\frac{6.63*10^(-34)}{\sqrt{2*9.1*10^(-31)*(2.176*10^(-20))}}`

`\lambda=3.33*10^(-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`

Answer 2

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

Step-by-step explanation:

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

`v_(n)=(2.2* 10^(6) m/s )/(n)`

Now the de Broglie wavelength can be calculated as,

`\lambda=(h)/(mv)`

For n=1 the wavelength is,

`\lambda=(6.626* 10^(-34)Js )/(9.1* 10^(-31)kg(2.2* 10^(6) m/s) )\\\lambda=0.33* 10^(-9)m\\ \lambda=0.33nm`

Therefore wavelength for n=1 orbit is 0.33 nm.

For n=4 the wavelength is,

`\lambda=(6.626* 10^(-34)Js )/(9.1* 10^(-31)kg((2.2* 10^(6) m/s)/(4) ) )\\\lambda=1.32* 10^(-9)m\\ \lambda=1.32nm`

Therefore wavelength for n=4 orbit is 1.32 nm.

For n=4 the wavelength is,

`\lambda=(6.626* 10^(-34)Js )/(9.1* 10^(-31)kg((2.2* 10^(6) m/s)/(4) ) )\\\lambda=3.3* 10^(-9)m\\ \lambda=3.3nm`

Therefore wavelength for n=10 orbit is 3.3 nm.