Question

A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be released by an electron in a mercury atom to produce a photon of this light?

a) 4.56 × 10^−19 J
b) 2.89 × 10^−19 J
c) 4.56 × 10^−18 J
d) 2.89 × 10^−18 J

Answer 1

The energy released by an electron to produce a 435.8 nm photon is calculated using Planck's equation and the given wavelength. Upon computation, the energy is found to be 4.56 × 10⁻¹⁹ J, which is option a).

Step-by-step explanation:

To calculate the amount of energy released by an electron in a mercury atom that produces a 435.8 nm photon of light, we can use the equation E = hc/λ, where h is Planck's constant (6.626 × 10⁻³⁴ J·s), c is the speed of light (3.00 × 10⁸ m/s), and λ is the wavelength of the photon (435.8 nm or 4.358 × 10⁻⁷ m).

First, we convert the wavelength into meters by multiplying it by 10⁻⁹, since 1 nm = 1 × 10⁻⁹ m.

Next, we calculate the energy (E) by plugging the values into the equation:

E = (6.626 × 10⁻³⁴ J·s × 3.00 × 10⁸ m/s) / 4.358 × 10⁻⁷ m = 4.56 × 10⁻¹⁹ J

Therefore, the correct amount of energy released by an electron to produce a 435.8 nm photon of light is 4.56 × 10⁻¹⁹ J, which corresponds to option a).