Question

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.

Answer 1

Answer:

$\large \boxed{4.578 \times 10^{-19} \text{ J}}$

Explanation:

The formula for the energy of a photon is

E = hc/λ

If λ = 433.9 nm, then  

$\begin{array}{rcl}E & = & \dfrac{6.626 \times 10^{-34} \text{ J \cdot s} \times 2.998 \times 10^{8}\text{ s}} {433.9 \times 10^{-9} \text{ m}}\\\\& = & 4.578 \times 10^{-19} \text{ J}\\\end{array}\\\text{The energy of one photon is \large \boxed{\mathbf{4.578 \times 10^{-19}} \textbf{ J}} }$

Answer 2

Answer:

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