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Question
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
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Solution
Step 1: Find the energy associated with the excited electron.
Given, energy of the electron beam used to bombard gaseous hydrogen =12.5 eV
We know, energy of the gaseous hydrogen atom in its ground state =−13.6 eV
After bombardment, the energy of the gaseous hydrogen atom becomes,
=−13.6+12.5=−1.1 eV
Step 2: Find the possible transitions.
Formula used: En=−13.6n2eV
As we know, energy of hydrogen atom in 𝑛^𝑡ℎ excited state, En==−13.6n2eV
So, −1.1=−13.6n2
n=3.5
As orbital number is integer value.
Therefore, n = 3
So, the electron beam excited the electron from n = 1 to n = 3.
During de-excitation the possible transition From n = 3 to n = 1(Lyman series)
From n = 2 to n = 1(Lyman series)
From n = 3 to n = 2 (Balmer series)
Step 3: Find wavelength associated with n = 3 to n =1transition.
Formula used: 1λ=Rc(1(n1)2−1(n2)2)
Step 3: Find wavelength associated with n = 3 to n =1 transition.
Formula used: 1λ=Rc(1(n1)2−1(n2)2)
When the electron undergoes transition from n = 3 to n = 1
Let wavelength associated with this transition be λ1.
1λ=Rc(1(n1)2−1n2)2)
Here, Rc = Rydberg constant =1.097×107m−1
1λ1=1.097×107(191)2−1(3)2)
λ1=1.0255×10−7m
λ1=103 nm
This radiation corresponds to the Lyman series of the hydrogen spectrum.
Step 4: Find wavelength associated with n = 2 to n =1 transition.
Formula used : 1λ=Rc(1(n1)2−1(n2)2)
When the electron undergoes transition from n = 2 to n = 1
Let wavelength associated with this transition be λ2.
1λ2=Rc(1(n1)2−1(n2)2)
Here, RC = Rydberg constant =1.097×107m−1
1λ2=1.097×107(1(11)2−1(22)2)
λ2=1.2154×10−7m
λ2=122 nm
This radiation corresponds to the Lyman series of the hydrogen spectrum.
Step 5: Find wavelength associated with n = 3 to n = 2 transition.
Formula used: (\dfrac{1}{\lambda} = R_c \left( \dfrac{1}{(n_1)^2} - \dfrac{1}{(n_2)^2}\right)\)
When the electron undergoes transition from n =3 to n =2
Let wavelength associated with this transition be λ3.
1λ3=Rc(1(n1)2−1(n2)2)
Here, Rc = Rydberg constant
=1.097×107m−1
1λ3=1.097×107(1(2)2−1(3)2)
λ3=6.57×10−7m
λ3=657 nm
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Final answer: Lyman series 103 nm, 122 nm; Balmer series 657 nm
Given, energy of the electron beam used to bombard gaseous hydrogen =12.5 eV
We know, energy of the gaseous hydrogen atom in its ground state =−13.6 eV
After bombardment, the energy of the gaseous hydrogen atom becomes,
=−13.6+12.5=−1.1 eV
Step 2: Find the possible transitions.
Formula used: En=−13.6n2eV
As we know, energy of hydrogen atom in 𝑛^𝑡ℎ excited state, En==−13.6n2eV
So, −1.1=−13.6n2
n=3.5
As orbital number is integer value.
Therefore, n = 3
So, the electron beam excited the electron from n = 1 to n = 3.
During de-excitation the possible transition From n = 3 to n = 1(Lyman series)
From n = 2 to n = 1(Lyman series)
From n = 3 to n = 2 (Balmer series)
Step 3: Find wavelength associated with n = 3 to n =1transition.
Formula used: 1λ=Rc(1(n1)2−1(n2)2)
Step 3: Find wavelength associated with n = 3 to n =1 transition.
Formula used: 1λ=Rc(1(n1)2−1(n2)2)
When the electron undergoes transition from n = 3 to n = 1
Let wavelength associated with this transition be λ1.
1λ=Rc(1(n1)2−1n2)2)
Here, Rc = Rydberg constant =1.097×107m−1
1λ1=1.097×107(191)2−1(3)2)
λ1=1.0255×10−7m
λ1=103 nm
This radiation corresponds to the Lyman series of the hydrogen spectrum.
Step 4: Find wavelength associated with n = 2 to n =1 transition.
Formula used : 1λ=Rc(1(n1)2−1(n2)2)
When the electron undergoes transition from n = 2 to n = 1
Let wavelength associated with this transition be λ2.
1λ2=Rc(1(n1)2−1(n2)2)
Here, RC = Rydberg constant =1.097×107m−1
1λ2=1.097×107(1(11)2−1(22)2)
λ2=1.2154×10−7m
λ2=122 nm
This radiation corresponds to the Lyman series of the hydrogen spectrum.
Step 5: Find wavelength associated with n = 3 to n = 2 transition.
Formula used: (\dfrac{1}{\lambda} = R_c \left( \dfrac{1}{(n_1)^2} - \dfrac{1}{(n_2)^2}\right)\)
When the electron undergoes transition from n =3 to n =2
Let wavelength associated with this transition be λ3.
1λ3=Rc(1(n1)2−1(n2)2)
Here, Rc = Rydberg constant
=1.097×107m−1
1λ3=1.097×107(1(2)2−1(3)2)
λ3=6.57×10−7m
λ3=657 nm
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Final answer: Lyman series 103 nm, 122 nm; Balmer series 657 nm
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