An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. if the photon emitted has a wavelength of 487 nm, what is the value of ni?

initial energy state is n = 4

Explanation:

For hydrogen atom when electron makes transition from higher energy state to lower energy state then the photon is released of energy which equal to the energy difference of two states

so here we can say

so here we have energy at nth excited state of electron is given as

here z = 1 for hydrogen

so now we have

here we know that

now plug in all values in it

by solving above equation we got

n = 4

b. lifting dumbbell weight at 20-degree angle from the body after a 45-degree angle

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg