Of the following transitions in the bohr hydrogen atom, the transition results in the absorption of the highest-energy photon. n = 2 → n = 1 n = 2 → n = 3 n = 5 → n = 6 n = 7 → n = 5

The n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

Explanation:

Formula used for the radius of the orbit will be,

where,

= energy of orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

Energy of the second orbit in H atom .

Energy of the third orbit in H atom .

Energy of the fifth orbit in H atom .

Energy of the sixth orbit in H atom .

Energy of the seventh orbit in H atom .

During an absorption of energy electron jumps from lower state to higher state.So,  absorption will take place in :

1) n = 2 → n = 3

2) n=  5 → n = 6

Energy absorbed when: n = 2 → n = 3

Energy absorbed when: n = 5 → n = 6

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

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