Question
Draw the current-voltage characteristics for the device shown in figure. between the terminals A and B.
✓
Answer
- If a battery is connected between terminals A and B, with positive terminal connected to point A and negative terminal connected to point B, then the diode will get forward biassed by the applied voltage. So, the current voltage graph for this circuit will be the same as that of the characteristic curves of a forward-biassed diode.
- If a battery is connected between terminals A and B, with positive terminal connected to point A and negative terminal connected to point B, then the upper diode will get forward biassed and the lower diode will get reverse biassed by the applied voltage. So, this lower branch can be replaced by an open circuit; hence, the current flow through this branch will be zero. The current flows only through the upper diode, so the circuit on simplification will become identical to the circuit in part (a). Hence, the current voltage graph for this circuit will be the same as that of the characteristic curves of a forward-biassed diode.
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Consider the fission of $^{238}_{92}\text{U}$ by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are $^{140}_{58}\text{Ce}$ and $^{99}_{44}\text{Ru}.$ Calculate Q for this fission process.
The relevant atomic and particle masses are:
$\text{m}(^{238}_{92}\text{U})=238.05079\text{ u}$
$\text{m}(^{140}_{58}\text{Ce})=139.90543\text{ u}$
$\text{m}(^{99}_{44}\text{Ru})=98.90594\text{ u}$
→The relevant atomic and particle masses are:
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$\text{m}(^{140}_{58}\text{Ce})=139.90543\text{ u}$
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