How much work is done to move 1 uc of charge from the negative terminal to the positive terminal of a 1.5v battery?

1.5 x 10⁻⁶ J

Explanation:

q = magnitude of charge on the charge = 1 x 10⁻⁶ C

ΔV = Potential difference between positive and negative terminals of battery = 1.5 Volts

W = Amount of work done to move the charge from negative terminal to the positive terminal

Work done is given as

W = q ΔV

W = (1 x 10⁻⁶) (1.5)

W = 1.5 x 10⁻⁶ J

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.