Question

Calculate the frequency and energy of a photon of radiation having wavelength 6000 angstrom

Answer 1

Answer:

Answer:lamda=6000A

Frequency(f)=velocity(c)/lamda

F=speed of light/lamba

F=3×10^8/6000×10^-10

=0.5×10^15hz

F=5×10^14hz

Energy=hf

h=Planck constant

=6.63×10^-34×5×10^-14

E=3.315×10^-47J

Answer 2

Answer:

The frequency and energy of a photon are 5×10¹⁴Hz and 3.3× 10⁻¹⁹J respectively.

Explanation:

The energy of the photon is given as

$E=h\nu$           (1)

E=energy of the photon

h=planks constant

ν=frequency of the photon

The energy of the photon is also given as,

$E=\frac{hc}{\lambda}$            (2)

c=speed of light=3×10⁸m/s

λ=wavelength of photon=6000 A⁰=6 ×10⁻⁷m

From equations (1) and (2) we have;

$h\nu=\frac{hc}{\lambda}$

$\nu=\frac{c}{\lambda}$         (3)

By substituting the values of c and λ in equation (3) we get;

$\nu=\frac{3\times 10^8}{6\times 10^-^7}=0.5\times 10^1^5=5\times 10^1^4Hz$

By substituting the value of ν in equation (1) we get;

$E=6.6\times 10^-^3^4\times 5 \times 10^1^4=33\times 10^-^2^0=3.3\times 10^-^1^9J$

Hence, the frequency and energy of a photon of radiation having wavelength 6000 angstrom is 5×10¹⁴Hz and 3.3× 10⁻¹⁹J respectively.