The electric potential at the surface of an atomic nucleus (z = 50) of radius 9 × 10-15 m is ______________.
8 × 106 volt
Let's break down how to find the electric potential at the surface of an atomic nucleus. The problem provides the atomic number (z) and the radius of the nucleus (r). We need to calculate the electric potential at the surface.
Understanding Electric Potential at the Nucleus Surface
An atomic nucleus contains protons, which are positively charged particles. The number of protons is equal to the atomic number (z). The total charge of the nucleus is therefore the number of protons multiplied by the charge of a single proton.
For calculations involving electric potential far from the nucleus compared to its size, or even at its surface treating it as a uniform sphere, we can often approximate the nucleus's charge as being concentrated at its center. This is the point charge approximation.
Key Concepts and Formula
- Atomic Number (z): The number of protons in the nucleus. In this case, z = 50.
- Charge of a Proton (e): The fundamental unit of charge, approximately \(1.6 \times 10^{-19}\) C.
- Total Charge of the Nucleus (Q): \(Q = z \times e\).
- Radius of the Nucleus (r): The distance from the center to the surface, given as \(9 \times 10^{-15}\) m.
- Electric Potential (V): The electric potential at a distance 'r' from a point charge 'Q' in vacuum is given by the formula: \[V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}\] where \(\frac{1}{4\pi\epsilon_0}\) is Coulomb's constant, approximately equal to \(9 \times 10^9\) Nm\(^2\)/C\(^2\).
Step-by-Step Calculation of Electric Potential
First, we need to calculate the total charge of the nucleus (Q).
Step 1: Calculate the total charge of the nucleus (Q)
Given z = 50 and \(e = 1.6 \times 10^{-19}\) C.
\(Q = z \times e\)
\(Q = 50 \times (1.6 \times 10^{-19}\) C)
\(Q = (50 \times 1.6) \times 10^{-19}\) C
\(Q = 80 \times 10^{-19}\) C
\(Q = 8.0 \times 10^{-18}\) C
Next, we use the formula for electric potential with the calculated charge and the given radius.
Step 2: Calculate the electric potential (V) at the surface
Given \(Q = 8.0 \times 10^{-18}\) C, \(r = 9 \times 10^{-15}\) m, and \(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\) Nm\(^2\)/C\(^2\).
\[V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}\]
\[V = (9 \times 10^9) \times \frac{8.0 \times 10^{-18}}{9 \times 10^{-15}}\]
Now, let's separate the numerical and exponential terms:
\[V = \left(9 \times \frac{8.0}{9}\right) \times \left(\frac{10^9 \times 10^{-18}}{10^{-15}}\right)\]
\[V = (8.0) \times \left(10^{9 - 18 - (-15)}\right)\]
\[V = 8.0 \times \left(10^{9 - 18 + 15}\right)\]
\[V = 8.0 \times \left(10^{-9 + 15}\right)\]
\[V = 8.0 \times 10^6\text{ V}\]
The electric potential at the surface of the atomic nucleus is \(8 \times 10^6\) Volts.
Comparing with Options
Let's compare our calculated value with the given options:
- Option 1: 80 volt
- Option 2: \(8 \times 10^6\) volt
- Option 3: 9 volt
- Option 4: \(9 \times 10^5\) volt
Our calculated value, \(8 \times 10^6\) V, matches Option 2.
Summary of Calculation
The electric potential at the surface of the atomic nucleus with z=50 and radius \(9 \times 10^{-15}\) m is calculated using the total charge (z times proton charge) and the radius in the potential formula. The result is \(8 \times 10^6\) volts.
Revision Table: Key Physics Concepts
| Concept | Description | Formula |
|---|---|---|
| Electric Potential (V) | Potential energy per unit charge at a point in an electric field. | \(V = \frac{PE}{q_0}\) or \(V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}\) (for point charge) |
| Electric Charge of Nucleus | Sum of charges of all protons (z \(\times\) e). | \(Q = z \times e\) |
| Coulomb's Constant | Proportionality constant in Coulomb's law and electric potential formula. | \(\frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9\) Nm\(^2\)/C\(^2\) |
Additional Information: Electric Potential and Fields
Electric potential is a scalar quantity, unlike the electric field which is a vector quantity. The electric potential is related to the electric field. For a point charge, the electric field points radially outward (for positive charge) and its magnitude is \(E = \frac{1}{4\pi\epsilon_0} \frac{|Q|}{r^2}\). The electric potential decreases as you move away from a positive charge.
The electric potential at the surface of a uniformly charged sphere of radius R and total charge Q is the same as the potential at a distance R from a point charge Q located at the center, provided the point is outside or on the surface of the sphere. Inside the uniformly charged sphere, the potential behavior is different, but at the surface, the point charge model gives the correct result for potential.
Nuclear radii are typically in the femtometer (\(10^{-15}\) m) range, which is why the potentials at the surface are very high.
Important Questions from Electrostatics
Electric potential $V$, is a ____ field, and electric field intensity $E$, is a ______ field.
The dielectric constant of a vacuum is _____.
- What is the magnitude of the electric field at a distance $r$ from a point charge $Q$?
According to Gauss’s Law, the surface integral of the normal component of electric flux density D over a closed surface containing charge Q is:
What is the relative permittivity of slate dielectric?
