how many of its photons are required to ionize atomic hydrogen? Einstein quote for Lenard’s measurement of the “largest effective wavelength for air”=190 nm,, For this problem, take the energy of the ground state of the hydrogen atom to be −0.99946650834R∞hc
Well, you've stumbled upon quite the cosmic conundrum! To ionize atomic hydrogen, we need to "energize" those electrons and kick them out of their comfy orbits. It turns out that each hydrogen atom requires an energy of about 13.6 electron volts (eV) to get the job done.
Now, let's convert this energy into good ol' photons. The energy of a single photon is given by Einstein's equation E = hv, where E is the energy, h is Planck's constant (approximately 6.63 x 10^-34 J.s), and v is the frequency of the photon.
For simplicity's sake, let's consider the largest effective wavelength for air given by Lenard - 190 nm. Now, wavelength (λ) and frequency (v) are inversely related by the speed of light (c), which is about 3 x 10^8 meters per second. So, v = c/λ.
Substituting the values, we find v ≈ (3 x 10^8 m/s) / (190 x 10^-9 m) ≈ 1.58 x 10^15 Hz.
Now let's find the energy of a single photon using the equation E = hv. Plugging in the frequency, we get E ≈ (6.63 x 10^-34 J.s) x (1.58 x 10^15 Hz) ≈ 1.05 x 10^-18 J.
Finally, let's convert this energy into electron volts (eV) because, hey, why not? We know that 1 eV is equal to approximately 1.60 x 10^-19 J. So, E ≈ (1.05 x 10^-18 J) / (1.60 x 10^-19 J/eV) ≈ 6.56 eV.
Phew! Now, to ionize atomic hydrogen, we need 13.6 eV. Thus, approximately 13.6 eV / 6.56 eV ≈ 2.07 photons are required to boot that electron out of its orbital home.
But hey, keep in mind that this is just an approximation. The quantum world is full of surprises, much like a clown car at a circus!
To determine how many photons are required to ionize atomic hydrogen, we need to calculate the energy difference between the ground state and the ionization energy.
The ionization energy of atomic hydrogen is the energy required to completely remove an electron from the atom. In this case, we are given that the energy of the ground state is -0.99946650834R∞hc, where R∞ is the Rydberg constant and hc is the product of Planck's constant (h) and the speed of light (c).
The ionization energy is the difference between the energy of the ionized state (where the electron is completely removed) and the energy of the ground state, which can be calculated as follows:
Ionization Energy = 0 - (-0.99946650834R∞hc)
= 0.99946650834R∞hc
Now, we need to calculate the energy of each photon. According to Einstein's photoelectric equation:
Energy of a photon = Planck's constant (h) x frequency (ν)
To find the frequency, we can use the equation:
Speed of light (c) = wavelength (λ) x frequency (ν)
Rearranging the equation, we have:
Frequency (ν) = Speed of light (c) / wavelength (λ)
Plugging in the given wavelength of 190 nm (or 190 x 10^-9 m), we can calculate the frequency:
Frequency (ν) = (3 x 10^8 m/s) / (190 x 10^-9 m)
= 1.57894736842 x 10^15 Hz
Finally, we can calculate the energy of each photon using Planck's constant (h):
Energy of a photon = Planck's constant (h) x frequency (ν)
= (6.62607015 x 10^-34 J*s) x (1.57894736842 x 10^15 Hz)
Now we can plug in the values and perform the calculation:
Energy of a photon ≈ 1.04467956 x 10^-18 J
To find how many photons are required to ionize atomic hydrogen, we divide the ionization energy by the energy of each photon:
Number of photons = Ionization Energy / Energy of a photon
≈ 0.99946650834R∞hc / (1.04467956 x 10^-18 J)
The final result gives the approximate number of photons required to ionize atomic hydrogen.
To determine the number of photons required to ionize atomic hydrogen, we need to calculate the energy required to ionize the hydrogen atom and then divide it by the energy of a single photon.
The energy required to ionize the hydrogen atom can be obtained using the formula:
E_ionization = E_final - E_initial
where E_final represents the energy of the ionized hydrogen atom and E_initial represents the energy of the ground state of hydrogen. According to the information provided, E_initial = -0.99946650834R∞hc, where R∞ is the Rydberg constant (approximately 1.0973731568508 x 10^7 per meter) and hc is the product of Planck's constant (approximately 6.62607004 x 10^-34 Joule-seconds) and the speed of light (approximately 2.998 x 10^8 meters per second).
Now, let's calculate E_initial:
E_initial = -0.99946650834 * R∞ * hc
Next, we need to calculate the energy of a single photon. The energy of a photon can be determined using the equation:
E_photon = h * c / λ
where h is Planck's constant and c is the speed of light (as mentioned above), and λ is the wavelength of the photon. The given information states that Lenard's measurement for the largest effective wavelength for air is 190 nm, which is equivalent to 190 x 10^-9 meters.
Now, let's calculate E_photon:
E_photon = h * c / λ
E_photon = (6.62607004 x 10^-34) * (2.998 x 10^8) / (190 x 10^-9)
Once we have both E_initial and E_photon, we can calculate the number of photons required to ionize atomic hydrogen by dividing E_initial by E_photon:
Number of photons = E_initial / E_photon
Please note that the exact value of E_photon is necessary for the calculation, which requires precise constants. The approximate values provided here are rounded for simplicity.
I hope this explanation helps!