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Hydrogen atom emits blue light when it c...

Hydrogen atom emits blue light when it changes from `n = 4` energy level to the `n = 2` level. Which colour of light would te atom emit when it changes from the `n = 5` level to the `n = 2` level ?

A

Red

B

Yellow

C

Green

D

Violet

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AI Generated Solution

The correct Answer is:
To determine the color of light emitted when a hydrogen atom transitions from the n=5 energy level to the n=2 energy level, we can follow these steps: ### Step 1: Understand the Energy Levels The hydrogen atom has discrete energy levels denoted by quantum numbers n=1, n=2, n=3, n=4, and n=5. The energy of these levels increases as n increases. ### Step 2: Calculate the Energy Difference The energy difference between two levels can be calculated using the formula for the energy of a level in a hydrogen atom: \[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \] Thus, the energy difference (ΔE) between the n=5 and n=2 levels is: \[ \Delta E = E_2 - E_5 = -\frac{13.6}{2^2} - \left(-\frac{13.6}{5^2}\right) \] \[ \Delta E = -\frac{13.6}{4} + \frac{13.6}{25} \] \[ \Delta E = -3.4 + 0.544 = -2.856 \, \text{eV} \] ### Step 3: Determine the Wavelength of Emitted Light Using the energy-wavelength relationship: \[ E = \frac{hc}{\lambda} \] Where: - \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{Js}) \) - \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \) We can rearrange this to find the wavelength: \[ \lambda = \frac{hc}{E} \] Substituting the values: \[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{2.856 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV}} \] \[ \lambda \approx \frac{1.987 \times 10^{-25}}{4.5696 \times 10^{-19}} \] \[ \lambda \approx 4.35 \times 10^{-7} \, \text{m} = 435 \, \text{nm} \] ### Step 4: Identify the Color of Light The wavelength of approximately 435 nm falls within the blue-violet region of the visible spectrum. Therefore, the color of light emitted when the hydrogen atom transitions from n=5 to n=2 is violet. ### Final Answer The hydrogen atom emits violet light when it changes from the n=5 level to the n=2 level. ---

To determine the color of light emitted when a hydrogen atom transitions from the n=5 energy level to the n=2 energy level, we can follow these steps: ### Step 1: Understand the Energy Levels The hydrogen atom has discrete energy levels denoted by quantum numbers n=1, n=2, n=3, n=4, and n=5. The energy of these levels increases as n increases. ### Step 2: Calculate the Energy Difference The energy difference between two levels can be calculated using the formula for the energy of a level in a hydrogen atom: \[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \] ...
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