The blue line of the hydrogen emission spectrum has a wavelength of 433.9 nm. a hydrogen emission spectrum has a violet, a blue, a teal, and a red line. calculate the energy of one photon of this light.

The energy of one photon of this light is 4.568 × 10∧ -19 J  

Explanation:

Given data:

wavelength = 433.9 nm or 433.9 × 10∧-9 m

The value of speed of light is 3× 10∧8 m/s and plancks constant is 6.626 × 10∧-34 m²Kg/s

Formula:

E= hc/λ

E= energy of photon

h= plancks constant

c= speed of light

λ= wavelength of given light

Solution:

E= hc / λ

E = 6.626 × 10∧-34 m²Kg/s × 3× 10∧8 m/s / 433.9 × 10∧-9 m

E = 19.878 × 10∧-26 m²Kg/s² / 433.9 × 10∧-9 m

E = 0.04568 × 10∧ -17 J            (Kg m²/s² =J)

E = 4.568 × 10∧ -19 J  

Explanation:

The formula for the energy of a photon is

E = hc/λ

If λ = 433.9 nm, then  

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

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