Answer:
The energy associated with this transition is 3.04 × 10⁻¹⁹ J.
Explanation:
An electron absorbs radiation when it goes from level 2 to level 3. The wavelength associated to this transition can be calculated using Rydberg equation.
[tex]\frac{1}{\lambda } =R_{H}(\frac{1}{n_{1}^{2} }-\frac{1}{n_{2}^{2} } )[/tex]
where,
λ is the wavelength of the radiation
RH is the Rydberg constant for Hydrogen (1.10 × 10⁷ m⁻¹)
n₁ and n₂ are the levels (being n₁ < n₂)
In this case,
[tex]\frac{1}{\lambda } =1.10 \times 10^{7} m^{-1} .(\frac{1}{2^{2} }-\frac{1}{3^{2} } )\\\lambda = 6.54 \times 10^{-7} m[/tex]
We can calculate the energy associated to this radiation using Planck-Einstein equation.
E = h . ν = h . c / λ
where,
h is the Planck's constant (6.63 × 10⁻³⁴ J.s)
c is the speed of light (3.00 × 10⁸ m/s)
Then,
[tex]E=h.\frac{c}{\lambda } =6.63 \times 10^{-34} J.s .\frac{3.00 \times 10^{8}m/s}{6.54 \times 10^{-7}m} =3.04 \times 10^{-19}J[/tex]
Answer:
[tex]3.03x10^-19~J[/tex]
Explanation:
We have to use the equation for the hydrogen atom:
Δ
E=[tex]-2.18 x10^-18~J~(\frac{1}{n^2_f}-\frac{1}{n^2_i})[/tex]
For this case we go from n=2 to n=3, so: [tex]n^2_f=3[/tex] and [tex]n^2_i=2[/tex].
Now we have to plug the values into the equation:
ΔE=[tex]-2.18 x10^-18~J~(\frac{1}{3^2}-\frac{1}{2^2})[/tex]
ΔE= [tex]3.03x10^-19~J[/tex]
If the sign is positive we have to give energy to the system. If we have to give energy the system would consume energy. If we have consumption of energy we will have absorption of light.
