The electric potential at the surface of an atomic nucleus (Z=50) of radius 9×10^(-13) cm is α×10^6 V. What is the value of α? (Charge of proton 1.6×10^(-19) C)

Correct answer: 8. The electric potential at the surface of an atomic nucleus can be written as V=kQ/R=(k·Ze)/R ...1 From equation (1), it follows that V = (9×10^9×50×1.6×10^(-19))/(9×10^(-13)×10^(-2)) V = 8×10^6 V Hence, α=8.

Q74 / 90 · JEE Main 2024 (27 Jan Shift 2) · Physics

The electric potential at the surface of an atomic nucleus $(Z = 50)$ of radius $9 \times 10^{- 13} cm$ is $α \times 10^{6} V$ . What is the value of $α$ ?
(Charge of proton $1.6 \times 10^{- 19} C$ )

Correct answer: 8

Solution

The electric potential at the surface of an atomic nucleus can be written as

$V = \frac{k Q}{R} = \frac{k \cdot (Z e)}{R} . . . (1)$

From equation (1), it follows that

$\begin{array}{rcl} V & = & \frac{9 \times 10^{9} \times 50 \times 1.6 \times 10^{- 19}}{9 \times 10^{- 13} \times 10^{- 2}} V \\ = & 8 \times 10^{6} V \end{array}$

Hence, $α = 8$ .

Hence, the correct answer is 8.

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The electric potential at the surface of an atomic nucleus (Z=50) of radius 9×10-13 cm is α×106 V. What is the value of α?(Charge of proton 1.6×10-19 C)

Solution

Relevant Questions

The electric potential at the surface of an atomic nucleus $(Z = 50)$ of radius $9 \times 10^{- 13} cm$ is $α \times 10^{6} V$ . What is the value of $α$ ?
(Charge of proton $1.6 \times 10^{- 19} C$ )