Some instruments differentiate individual quanta of electromagnetic radiation based on their energies. assume such an instrument has been adjusted to detect quanta that have 3.50× 10–16 j of energy. what is the wavelength of the detected radiation? give your answer in nanometers.

5.68×10¯¹⁰ m

Explanation:

From the question given above, the following data were obtained:

Energy (E) = 3.50×10¯¹⁶ J

Wavelength (λ) =?

Energy (E) and wavelength (λ) are related according to the following formula:

E = hv/λ

Where

E => is the energy

h => is the Planck's constant

v => is the velocity of electromagnetic radiation

λ => is the wavelength of the radiation.

With the above formula, we can obtain the wavelength of the radiation as follow:

Energy (E) = 3.50×10¯¹⁶ J

Planck's constant (h) = 6.63×10¯³⁴ Js

Velocity (v) = 3×10⁸ m/s

Wavelength (λ) =?

E = hv/λ

3.50×10¯¹⁶ = 6.63×10¯³⁴ × 3×10⁸ / λ

3.50×10¯¹⁶ = 1.989×10¯²⁵/ λ

Cross multiply

3.50×10¯¹⁶ × λ = 1.989×10¯²⁵

Divide both side by 3.50×10¯¹⁶

λ = 1.989×10¯²⁵ / 3.50×10¯¹⁶

λ = 5.68×10¯¹⁰ m

Thus, the wavelength of the radiation is

5.68×10¯¹⁰ m

1.solid

2.carbon dioxide

3.helium

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).