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An atom requires 6.24 × 10−19J to remove its furthest electron in the gas phase. What is the maximum wavelength of electromagnetic radiation that will result in ionization? Report your numerical answer in units of nm. Use significant figures.

Respuesta :

johnkellock
Solution:
1. Determine the frequency:

E=hv
6.24 x 10-19 J = (6.626 x 10-32 Js) x (v)
v = 9.417 x 10^14 s-1

2. Determine the wavelength:

λv = c
(λ) x (9.417 x 10^14 s-1) = 3.00 x 10^8 m/s
λ = 3.18 x 10-7 m

Wavelength (λ) = 318nm

onyebuchinnaji

The maximum wavelength of electromagnetic radiation that will result in ionization is 318.6 nm.

The given parameters;

  • energy required by the atom, E = 6.24 x 10⁻¹⁹ J

The maximum wavelength of electromagnetic radiation that will result in ionization is calculated as follows;

E = hf

[tex]E = h\frac{c}{\lambda}[/tex]

where;

  • f is the frequency of the photon
  • c is the speed of light
  • λ is the wavelength

The maximum wavelength is calculated as;

[tex]\lambda = \frac{hc}{E} \\\\\lambda = \frac{(6.626\times 10^{-34}) \times (3\times 10^8)}{6.24\times 10^{-19}} \\\\\lambda = 3.186 \times 10^{-7} \ m\\\\\lambda = 318.6 \times 10^{-9} \ m\\\\\lambda = 318.6 \ nm[/tex]

Thus, the maximum wavelength of electromagnetic radiation that will result in ionization is 318.6 nm.

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