Energy Level Transfer of Hydrogen's electron. For each line, determine the prinicpal energy level to which hydrogen's electron was excited. When the electron fell to the 2nd level, energy was released corresponding to each of the lines.
E= 2.179X10-18 (1/n2 lo- 1/n2 hi). The value for "E" comes from your previous calculation. nlo= (2nd Principal energy level). You are looking for n hi.
To determine the principal energy level to which hydrogen's electron was excited for each line, we can use the energy level formula:
E = 2.179 × 10^-18 (1/nlo^2 - 1/nhi^2)
Here, E represents the energy released when the electron falls to the second level (nlo). We need to find the value of nhi for each line.
Let's go through the lines one by one:
1. Line 1:
E = 2.179 × 10^-18 (1/nlo^2 - 1/nhi^2)
For the first line, nlo = 2nd level. We need to find nhi.
Plugging in the values we have:
E = 2.179 × 10^-18 (1/2^2 - 1/nhi^2)
Simplifying the equation:
E = 2.179 × 10^-18 (1/4 - 1/nhi^2)
Now, we can set this equal to the known value of E and solve for nhi:
2.179 × 10^-18 (1/4 - 1/nhi^2) = E
Solving for nhi, we get:
1/4 - 1/nhi^2 = E/(2.179 × 10^-18)
Next, we can rearrange the equation:
1/nhi^2 = (E/(2.179 × 10^-18)) + 1/4
Now, let's solve for nhi:
nhi^2 = 1 / ((E/(2.179 × 10^-18)) + 1/4)
Finally, taking the square root of both sides:
nhi = √(1 / ((E/(2.179 × 10^-18)) + 1/4))
Repeat this process for each line, substituting the corresponding value of E and solving for nhi using the above formula.
To determine the principal energy level to which hydrogen's electron was excited for each line, we can use the formula provided:
E = 2.179 × 10^(-18) * (1/n₂ lo - 1/n₂ hi)
In this equation, E represents the energy released when the electron fell to the 2nd energy level, n₂ lo represents the 2nd principal energy level, and n₂ hi represents the unknown principal energy level we are trying to find.
To solve for n₂ hi, we can rearrange the equation as follows:
1/n₂ hi = 1/n₂ lo - E / (2.179 × 10^(-18))
Now, let's substitute the known values. Since the electron fell to the 2nd energy level, n₂ lo = 2. The value of E should have been calculated previously in order to proceed.
1/n₂ hi = 1/2 - E / (2.179 × 10^(-18))
Finally, we can solve for n₂ hi by taking the reciprocal of both sides of the equation:
n₂ hi = 1 / (1/2 - E / (2.179 × 10^(-18)))
Simply plug in the value of E calculated from the previous calculation, and then evaluate the expression to find the principal energy level (n₂ hi) to which hydrogen's electron was excited for each line.