Question

A black light produces UV light with wavelengths below 400 nm. When the chemical quinine (which is in tonic water) is exposed to light with a wavelength of 383 nm, it becomes excited and then re-emits light at 450. nm.

Is the re-emitted light lower or higher energy than the light that was absorbed? Is it lower or higher frequency? Answer this question without a calculation.

What is the energy of a single photon of the UV light that was emitted?

  

  

Answer 1

The (invisible) ultraviolet light from the black light is absorbed by the quinine in the tonic water,

and this excites the quinine.

When the quinine becomes unexcited, it releases visible blue light that we see.

bLUE LIGHT WAVES ARE OF SHORTEST WAVELNEGTH AND HIGHEST ENERGY THEREFORE THE RE EMIITED LIGHT HAS HIGHER ENERGY.

UV LIGHT HAS HIGHER FREQUENCY (SHORTER WAVELENGTH) THAN VISIBLE LIGHT.

ENERGY OF PHOTON, E =h v

h is Plancks constant

v is frequency

And E is energy in joules

V OF UV LIGHT=

=6.626 * 10 ^ -34 * vWe have to find frequency

c =wavelegth * freaquency

Frequency=c/wavelength

c is speed of light

=3 * 10 ^8/400 * 10 ^-9

= 0.0075 * 10 ^17

Energy= 6.626 * 10 ^-34 *0.0075 * 10 ^17

=0.05 *10^-17 Joules