1. My laser pointer outputs 3 mW of green light at at 550nm. How many photons le

Question

1. My laser pointer outputs 3 mW of green light at at 550nm. How many photons leave the pointer each second? options: A) 8.3 E15 photons/sec B) 1.7 E24 photons/sec C) 7.1 E16 photons/sec D) 1 photon/sec E) 5.5 E18 photons/sec

2. What is the role of the phosphor coating in a low pressure mercury discharge lamp? options: A) To shield us from the dangerous UV light emitted by the mercury atoms B) Mercury atoms hit the phosphor molecules and then the phosphor molecules emit the white photons needed for white light. C) Phosphor molecules help to recycle mercury atoms and prolong the life of the bulb D) When excited phosphor molecules emit a continuous spectrum of wavelengths, enabling the bulb to produce light that we perceive as white. E) The different phosphor molecules in the coating produce a sufficient range of different visible wavelengths (when excited) to convince the human eye that it is seeing white light.

Explanation / Answer

The energy of a photon is given by the equation E = hc/, where E is the energy, Planck's constant h = 6.63 x 10^(-34) (with nasty units), c = 3.00 x 10^8 m/s (speed of light) and is the wavelength of the light (in metres).

So in your case, =  550x 10^(-9), so the energy of one photon is:

E = 3.616 * 10^(-19) Joules.

3 mW means 0.003 W, which is 0.003 Joules per second. So the number of photons released per second is 0.003 / (3.616 x 10^(-19)) = 8.296 x 10^15.

Hence the answer for the first question is A) 8.3 E15 photons/sec.

For the second part let me explain in a bit detail,

Since the very beginnings of mercury lighting it had been realised that a significant portion of the energy radiated by all mercury discharges is in the ultra-violet part of the spectrum. In the case of the low pressure discharge, more than half of the total energy supplied is radiated in the short-wave UV region at 253.7nm. High pressure lamps radiate about 10% of their energy in the long-wave UV region at 365.0nm, but an appreciable amount is also radiated at shorter wavelengths. In general lighting applications this UV is wasted since it is invisible to human eyes. But it was known from the outset that certain materials called phosphors exist, which are capable of fluorescing under ultra-violet radiation and generating light of other colours.

In the mid 1930's such materials were put to use in the development of the fluorescent tube, based on converting the plentiful shortwave UV radiation from a low pressure mercury discharge into visible light. The phosphor coating was applied to the inside of the glass tube in which the mercury discharge took place.

Hence option E) The different phosphor molecules in the coating produce a sufficient range of different visible wavelengths (when excited) to convince the human eye that it is seeing white light.

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