A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron $$= 1.6 \times 10^{-19}$$ C)

We are told that the particle carries the same charge as an electron, so its charge is

$$q = 1.6 \times 10^{-19}\ \text{C}.$$

Under only the magnetic field, the particle executes uniform circular motion of radius

$$r = 0.5\ \text{cm} = 0.5 \times 10^{-2}\ \text{m} = 5 \times 10^{-3}\ \text{m}.$$

The magnetic field magnitude is

$$B = 0.5\ \text{T}.$$

For a charged particle moving perpendicular to a uniform magnetic field, the magnetic force supplies the centripetal force. We first state this relationship:

Magnetic force: $$F_B = q v B.$$ Centripetal force: $$F_C = \dfrac{m v^{2}}{r}.$$

Because the motion is circular under the magnetic field alone, these two forces are equal:

$$q v B = \dfrac{m v^{2}}{r}.$$

We can cancel one factor of the speed $$v$$ from both sides to find an expression for $$v$$:

$$q B = \dfrac{m v}{r}\quad \Longrightarrow\quad v = \dfrac{q B r}{m}.$$

Next, an electric field of magnitude

$$E = 100\ \text{V m}^{-1}$$

is switched on simultaneously with the magnetic field. The problem states that, under the simultaneous action of $$ \mathbf{E} $$ and $$ \mathbf{B} $$, the particle now moves in a straight line. A straight-line trajectory in crossed, perpendicular electric and magnetic fields implies that the net force is zero. The electric force and magnetic force must therefore cancel:

Electric force: $$F_E = q E.$$ Magnetic force (already written): $$F_B = q v B.$$

For cancellation, we must have

$$q E = q v B.$$

Because $$q \neq 0$$, we divide both sides by $$q$$ to obtain the standard crossed-fields condition:

$$E = v B \quad\Longrightarrow\quad v = \dfrac{E}{B}.$$

We now have two independent expressions for the speed $$v$$:

$$v = \dfrac{q B r}{m}\qquad\text{and}\qquad v = \dfrac{E}{B}.$$

Since both describe the same speed, we equate them:

$$\dfrac{q B r}{m} = \dfrac{E}{B}.$$

Solving this equation for the mass $$m$$, we first cross-multiply:

$$q B r \, B = m E.$$

This simplifies to

$$m = \dfrac{q B^{2} r}{E}.$$

Now we substitute the numerical values, keeping all factors explicit:

$$m = \dfrac{(1.6 \times 10^{-19}\ \text{C})\,(0.5\ \text{T})^{2}\,(5 \times 10^{-3}\ \text{m})}{100\ \text{V m}^{-1}}.$$

First calculate the square of the magnetic field:

$$(0.5\ \text{T})^{2} = 0.25\ \text{T}^{2}.$$

Multiply $$q$$ and $$B^{2}$$:

$$1.6 \times 10^{-19} \times 0.25 = 0.4 \times 10^{-19} = 4.0 \times 10^{-20}.$$

Now include the radius $$r$$:

$$4.0 \times 10^{-20} \times 5 \times 10^{-3} = 20.0 \times 10^{-23} = 2.0 \times 10^{-22}.$$

Finally, divide by the electric field magnitude $$E = 100 = 1.0 \times 10^{2}$$:

$$m = \dfrac{2.0 \times 10^{-22}}{1.0 \times 10^{2}} = 2.0 \times 10^{-24}\ \text{kg}.$$

This value exactly matches option C.

Hence, the correct answer is Option C.