Question

You see a flame that has a yellow color, and an emission line on the spectroscope at 6.03 mm. You have a calibration curve of y= 117.0x - 133.4 where y is the wavelength in nm and x is the position on the spectroscope in mm. What is the energy of this electron as it relaxes and emits a photon?

a) 2.82 eV
b) 4.67 eV
c) 5.32 eV
d) 6.75 eV

Answer 1

The energy of the emitted photon can be calculated using the equation E = hc/λ. Plugging in the values, the energy of the photon is found to be 2.066 eV.

Step-by-step explanation:

The energy of a photon can be calculated using the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength.

To convert the given wavelength of 6.03 mm to nm, we multiply by 1000, resulting in 6030 nm.

Plugging in the values into the equation, we get E = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (6030 x 10^-9 m) = 3.309 x 10^-19 J.

To convert from joules to electron volts (eV), we divide by 1.602 x 10^-19 J/eV. Therefore, the energy of the emitted photon is 3.309 x 10^-19 J / (1.602 x 10^-19 J/eV) = 2.066 eV.

Therefore, the correct answer is 2.066 eV (option b).