Question

The allowed energies of a simple atom are 0.0 eV, 4.0 eV, and 6.0 eV. Part A What wavelength(s) appear(s) in the atom's emission spectrum

Answer 1

Answer:

3.1 × 10^- 7 m and 2.1 × 10^-7 m

Explanation:

First we must convert each value of energy to Joules by multiplying its value by 1.6 ×10^-19. After that, we can now obtain the wavelength from E= hc/λ

Where;

h= planks constant

c= speed of light

λ= wavelength of light

For 6.0ev;

E= 6.0 × 1.6 ×10^-19

E= 9.6 × 10^-19 J

From

E= hc/λ

λ= hc/E

λ= 6.6 × 10^-34 × 3 × 10^8/9.6 × 10^-19

λ= 2.1 × 10^-7 m

For 4.0 eV

4.0 × 1.6 × 10^-19 = 6.4 × 10^-19 J

E= hc/λ

λ= hc/E

λ= 6.6 × 10^-34 × 3 × 10^8/6.4 × 10^-19

λ= 3.1 × 10^- 7 m