One of the emission spectral lines for be3+ has a wavelength of 466.4 nm for an electronic transmission that begins in the state with n = 6. what is the principal quantum number of the lower-energy state corresponding to this emission?

The answer is -6.05 x 10 ^-20. In order to get this answer, you have to use the Rydberg formula of -R(1/n ^2).
 Rydberg constant is -2.178x10 ^ -18
n= 6
so the solution is
=-2.178 x10 ^ -18 (1/ 6^2)
=-6.05 x 10 ^ -20

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.

He sandals were out in the sun so the got hotter. which is why her feet felt hot.