An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity $$ 10^{6} m/s$$ . If the magnitude of the electric field between the plates is $$ 9.1\text{ } V/cm $$, then the vertical component of velocity of electron is (mass of electron $$=9.1\times 10^{31}kg $$ and charge of electron $$ =1.6\times 10^{-19}C) $$

The electron enters symmetrically between the plates, meaning it starts at the midpoint with an initial vertical velocity of zero. The initial velocity is purely horizontal, denoted as $$u_x$$. The horizontal component of velocity remains constant because there is no force in the horizontal direction. After emerging, the horizontal component is given as $$10^6 \, \text{m/s}$$, so $$u_x = 10^6 \, \text{m/s}$$.

The length of each plate is $$L = 10 \, \text{cm} = 0.1 \, \text{m}$$. The time taken to cross the electric field region is calculated using the horizontal motion:

$$t = \frac{L}{u_x} = \frac{0.1}{10^6} = 10^{-7} \, \text{s}$$

The electric field magnitude is $$E = 9.1 \, \text{V/cm}$$. Convert to SI units: $$1 \, \text{V/cm} = 100 \, \text{V/m}$$, so $$E = 9.1 \times 100 = 910 \, \text{V/m}$$.

The force on the electron in the vertical direction is due to the electric field. The charge of the electron is $$q = -1.6 \times 10^{-19} \, \text{C}$$, but the magnitude of the force is $$|F| = eE$$, where $$e = 1.6 \times 10^{-19} \, \text{C}$$. The mass of the electron is $$m = 9.1 \times 10^{-31} \, \text{kg}$$.

The acceleration in the vertical direction is given by Newton's second law:

$$a_y = \frac{|F|}{m} = \frac{eE}{m}$$

Substitute the values:

$$a_y = \frac{(1.6 \times 10^{-19}) \times 910}{9.1 \times 10^{-31}}$$

First, compute the numerator: $$1.6 \times 10^{-19} \times 910 = 1.6 \times 10^{-19} \times 9.1 \times 10^2 = 14.56 \times 10^{-17} = 1.456 \times 10^{-16}$$

Now, divide by the denominator:

$$a_y = \frac{1.456 \times 10^{-16}}{9.1 \times 10^{-31}} = \frac{1.456}{9.1} \times 10^{-16 - (-31)} = 0.16 \times 10^{15} = 1.6 \times 10^{14} \, \text{m/s}^2$$

The initial vertical velocity is $$u_y = 0$$. The vertical component of velocity when the electron emerges is given by the equation of motion:

$$v_y = u_y + a_y t = 0 + (1.6 \times 10^{14}) \times (10^{-7}) = 1.6 \times 10^{14} \times 10^{-7} = 1.6 \times 10^{7} \, \text{m/s}$$

Simplify: $$1.6 \times 10^7 \, \text{m/s} = 16 \times 10^6 \, \text{m/s}$$.

The vertical component of velocity is $$16 \times 10^6 \, \text{m/s}$$, which corresponds to option C.

Final Answer: C. $$16 \times 10^{6} \, \text{m/s}$$