The bright red line from the emission spectrum of hydrogen has a wavelength of 657 nm. What is the energy, in joules, of a single photon of this light? 4.35 x 10–40 J 3.03 x 10–28 J 3.03 x 10–19 J 4.56 x 1014 J

3.03 × 10^-19 J

Explanation:

The energy, E, of a photon of light can be calculated by using the formula:

E = hv

Where; E is energy in Joules,

h is Planck's constant {6.626×10^−34J}

v is frequency of light

For this question, the wavelength (λ) is given, not the frequency (v).

Hence, we use;

v= c/λ

Where v= frequency, c= speed of light (3×10^8m/s), λ= wavelength (657nm)

657nm = 657 × 10^-9m

v = 3×10^8/657 × 10^-9

v = 3/657 × 10^(8--9)

v = 0.00456 × 10^17

v = 4.56 × 10^14

Since frequency (v) = 4.56 × 10^14;

E = hv

E = 6.626×10^−34 × 4.56 × 10^14

E = 30.21 × 10^(-34+14)

E = 30.21 × 10^-20

E = 3.021 × 10^-19

Therefore, the energy of the photon of light is 3.02 × 10^-19 J

there all right good

explanation:

i have no clue hope you find the answer