Different isotopes of the same element emit light at slightly different wavelengths, the minimum number of slits is mathematically given as
N=1820slits
<h3>What minimum number of slits is required to resolve these two wavelengths in second-order?</h3>
Generally, the equation for the wave is mathematically given as

Where the chromatic resolving power (R) is defined by

R = nN,
Therefore


and


In conclusion, the minimum number of slits is required to resolve these two wavelengths in second-order

Therefore

N=1820slits
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Answer:
increasing rainfall. hope it help
Answer:
the difference is tyat eruptions of less gassy and more gassy is that the less gassy doesnt retain as much gas as the more gassy one and thus the eruption of the less gassy is less damage to the more gassy
Answer:
I think D
Explanation:
Ok, I'm not sure but it sounds right ish you should check a practice video or something. It might also be B or C but im pretty certain it isnt A just ask yourself is the student measuring it in newtons? Is that important in the process? What about if the student is considering the affect of mass is it important? Good luck srry if im not much of help! If this is like A SUPER IMPORTANT TEST OR SOMETHING RLLLLLLLY IMPORTANT just wait for another answer gl!
S and S²⁻ do not have the outer subshell fully filled with electrons.
Explanation:
We look at electronic configurations:
Ca 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² - the outer subshell 4s² is fully-filled with electrons
S 1s² 2s² 2p⁶ 3s² 3p⁴ - the outer subshell 3p⁴ is not fully-filled with electrons
Zn²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s⁰ - here the 4s subshell is higher in energy than 3d subshell so will consider 3d¹⁰ the out subshell which is fully-filled with electrons
S²⁻ 1s² 2s² 2p⁶ 3s² 3p² - the outer subshell 3p² is not fully-filled with electrons
Ca²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ - the outer subshell 3p⁶ is fully-filled with electrons
Learn more about:
electron configurations
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