The first ionization energy, E, of a boron atom is 1.33 aJ. What is the wavelength of light, in nm, that is just sufficient to ionize a boron atom? Values for constants can be found here.
E = hc/λ
1.33 aJ = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/λ
λ = 2.20 x 10^-7 m = 220 nm
To find the wavelength of light that is just sufficient to ionize a boron atom, we can use the equation:
E = hc/λ
Where:
- E is the ionization energy
- h is Planck's constant (provided in the constants)
- c is the speed of light (also provided in the constants)
- λ is the wavelength we want to find
Rearranging the equation to solve for λ, we get:
λ = hc/E
Now we can substitute the given values:
E = 1.33 aJ = 1.33 x 10^-18 J (converting aJ to J using the prefix conversion factor)
h = Planck's constant = 6.626 x 10^-34 J.s (provided in the constants)
c = speed of light = 2.998 x 10^8 m/s (provided in the constants)
Plugging these values into the equation:
λ = (6.626 x 10^-34 J.s)(2.998 x 10^8 m/s) / (1.33 x 10^-18 J)
λ = 1.489 x 10^-6 m
However, the question asks for the wavelength in nanometers (nm). To convert from meters to nanometers, we use the conversion factor:
1 meter = 10^9 nanometers
λ = 1.489 x 10^-6 m x (10^9 nm/1m)
λ = 1489 nm
Therefore, the wavelength of light that is just sufficient to ionize a boron atom is approximately 1489 nm.
To determine the wavelength of light that is just sufficient to ionize a boron atom, we can use the equation:
E = hc/λ
Where:
E is the ionization energy (in joules)
h is Planck's constant (6.62607015 × 10^-34 J·s)
c is the speed of light (2.998 × 10^8 m/s)
λ is the wavelength of light (in meters)
First, let's convert the ionization energy to joules from the given value of 1.33 aJ (attojoules), which is 1.33 × 10^-18 J.
E = 1.33 × 10^-18 J
Now, we can rearrange the equation to solve for λ:
λ = hc/E
Substituting the values:
λ = (6.62607015 × 10^-34 J·s) × (2.998 × 10^8 m/s) / (1.33 × 10^-18 J)
Calculating this, we find:
λ ≈ 1.498 × 10^-7 m
Finally, let's convert the wavelength from meters to nanometers:
λ = 1.498 × 10^-7 m × (10^9 nm/1 m)
λ ≈ 149.8 nm
Therefore, the wavelength of light that is just sufficient to ionize a boron atom is approximately 149.8 nm.