580 nm light shines on a double slit with d=0.000125 m. What is the angle of the third dark interference minimum (m=3)? (Remember, nano means 10^-9.)
(Unit=deg)

0.66 degrees

Explanation:

The computation of the angle of the third dark interference is shown below:

The condition of the minima is

Path difference = (2n +1) × ÷ 2

For third minima, n = 2

Now

xd ÷ D = (2 × 2 + 1) × ÷ 2

d tan Q_3 = 5 ÷ 2

tan Q_3 = 5 ÷ 2d

Q_3 = tan^-1 × (5 ÷2d)

= tan^-1 × (5 × 580 × 10^-9) ÷ (2 × 0.000125)

= 0.66 degrees

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases . this process does not change the mass number.

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as -decay. in this the mass number remains same.

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.