Photo AI

9. A laser emits light when electrons are stimulated to fall from a high energy level to a lower energy level - Scottish Highers Physics - Question 9 - 2019

Question icon

Question 9

9.-A-laser-emits-light-when-electrons-are-stimulated-to-fall-from-a-high-energy-level-to-a-lower-energy-level-Scottish Highers Physics-Question 9-2019.png

9. A laser emits light when electrons are stimulated to fall from a high energy level to a lower energy level. The diagram shows some of the energy levels involved. ... show full transcript

Worked Solution & Example Answer:9. A laser emits light when electrons are stimulated to fall from a high energy level to a lower energy level - Scottish Highers Physics - Question 9 - 2019

Step 1

Determine the wavelength of the photon emitted.

96%

114 rated

Answer

To find the wavelength of the photon emitted during the transition from E1 to E5, we first calculate the change in energy:

E=E1−E5=−2.976×10−18 J−(−3.290×10−18 J)E = E_1 - E_5 = -2.976 \times 10^{-18} \text{ J} - (-3.290 \times 10^{-18} \text{ J})

Calculating this gives:

E=3.00×10−18 JE = 3.00 \times 10^{-18} \text{ J}

Next, we use the relationship between energy and wavelength:

E=hcλE = \frac{hc}{\lambda}

Where:

  • hh (Planck's constant) = 6.63×10−34 Js6.63 \times 10^{-34} \text{ Js}
  • cc (speed of light) = 3.00×108 m/s3.00 \times 10^{8} \text{ m/s}

Rearranging for wavelength (λ\lambda) gives us:

λ=hcE\lambda = \frac{hc}{E}

Substituting the values:

λ=(6.63×10−34)(3.00×108)3.00×10−18≈6.63×10−7 m=663 nm\lambda = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{3.00 \times 10^{-18}} \approx 6.63 \times 10^{-7} \text{ m} = 663 \text{ nm}

Thus, the wavelength of the photon emitted is approximately 663 nm.

Step 2

Determine the power of the laser beam.

99%

104 rated

Answer

To find the power of the laser beam, we first calculate the area of the spot of light on the screen, which is circular:

A=πr2A = \pi r^2

Where the radius r=d2=8.00×10−42=4.00×10−4 mr = \frac{d}{2} = \frac{8.00 \times 10^{-4}}{2} = 4.00 \times 10^{-4} \text{ m}.

Thus:

A=π(4.00×10−4)2≈5.03×10−7 m2A = \pi (4.00 \times 10^{-4})^2 \approx 5.03 \times 10^{-7} \text{ m}^2

Next, using the irradiance (II) formula, we can find the power (PP):

I=PAightarrowP=I⋅AI = \frac{P}{A} ightarrow P = I \cdot A

Substituting the known values:

P=9950 W m−2⋅5.03×10−7 m2≈5.00×10−3 WP = 9950 \text{ W m}^{-2} \cdot 5.03 \times 10^{-7} \text{ m}^2 \approx 5.00 \times 10^{-3} \text{ W}

Thus, the power of the laser beam is approximately 0.005 W.

Step 3

Describe how this apparatus could be used to verify the inverse square law for a point source of light.

96%

101 rated

Answer

To verify the inverse square law for a point source of light using the apparatus:

  1. Setup the apparatus: Ensure that the light source, light sensor, and measuring stick are positioned correctly with the light source at a fixed distance from the sensor.

  2. Measure irradiance: Record the irradiance (I) at a certain distance (d) from the light source using the light sensor. Ensure accurate readings.

  3. Repeat measurements: Gradually increase the distance (d) by known increments, and at each new distance, record the irradiance. Repeat this for multiple distances to build a dataset.

  4. Calculate results: According to the inverse square law, irradiance varies inversely with the square of the distance from the source:

I∝1d2I \propto \frac{1}{d^2}
  1. Graph data: Plot a graph of I against rac{1}{d^2}. If the graph is a straight line (linear relationship), then it verifies the inverse square law.

In this manner, you can validate the relationship between irradiance and distance for a point source of light.

Join the Scottish Highers students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;