91影视

$\text{V=20V}$

$D=10\text{m}$

${\text{d=10}}^{\text{-5}}\text{m}$

${\left|{\text{蠄}}_{\text{1}}\right|}^{\text{2}}\text{鈥夆赌夆赌坝扁赌夆赌夆赌�100}$

${\left|{蠄}_2\right|}^2\text{鈥夆赌夆赌�}伪\text{鈥夆赌夆赌�}900$

We know that when a charged particle of charge q and mass m is accelerated with a potential, $\mathbf{\text{V}}$, the kinetic energy$\mathbf{\left(}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}{\mathbf{\text{mv}}}^{\mathbf{\text{2}}}\mathbf{\right)}$can be expressed as,

$\mathbf{\text{qV=}}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}{\mathbf{\text{mv}}}^{\mathbf{\text{2}}}$鈥︹赌︹赌︹赌︹赌︹赌︹赌︹赌�(1)

and that the velocity, v, is proportional to the wavelength using the de Broglie formula.

$\mathbf{\text{v=}}\frac{\mathbf{\text{p}}}{\mathbf{\text{m}}}\mathbf{\text{=}}\frac{\mathbf{\text{h}}}{\mathbf{\text{尘位}}}$鈥︹赌︹赌�..鈥︹赌︹赌︹赌�(2)

We need to eliminate v by using the equations (1) and (2) to find the wavelength$\text{位}$of the electron.

First, we must describe both equations in terms of ${\text{v}}^{\text{2}}$, such that

$\frac{\text{2qV}}{\text{m}}{\text{=v}}^{\text{2}}$

${\text{v}}^{\text{2}}\text{=}\frac{{\text{h}}^{\text{2}}}{{\text{m}}^{\text{2}}{\text{位}}^{\text{2}}}$

Then, we combine both equations.

$\frac{\text{2qV}}{\text{m}}\text{=}\frac{{\text{h}}^{\text{2}}}{{\text{m}}^{\text{2}}{\text{位}}^{\text{2}}}$

$\text{2qV=}\frac{{\text{h}}^{\text{2}}}{{\text{尘位}}^{\text{2}}}$

Lastly, we isolate$\text{位}$

$\text{位=}\frac{\text{h}}{\sqrt{\text{2qVm}}}$鈥︹赌︹赌︹赌︹赌︹赌︹赌︹赌�..(3)

Now, substitute the given parameters in equation (3) to obtain the wavelength of the electron,

$\begin{array}{c}\text{位}=\frac{\text{6.63}脳{\text{10}}^{\text{-34}}\text{鈥�}J鈰�s}{\sqrt{\text{2}脳(\text{1.6}脳{\text{10}}^{\text{-19}}\text{鈥�}C)脳\text{(20鈥塚)}脳(\text{9.1}脳{\text{10}}^{\text{-31}}\text{鈥�}kg)}}\\ =\text{2.75}脳{\text{10}}^{\text{-10}}\text{m}\end{array}$

Thus the wavelength of the electron is${\text{2.75脳10}}^{\text{-10}}\text{鈥尘}$

(a)

Using the constructive interference equation, we get,

$\text{尘位=诲sin(胃)}$

Using sine law, we get,

$\frac{\text{1}}{\text{2}}\text{位=诲}\frac{\text{y}}{\text{D}}$

Hence, substituting the given information in the above equation, we get,

$\begin{array}{c}\text{y}=\frac{\text{D}}{\text{2d}}\text{位}\\ =\frac{\text{10}}{\text{2}脳\left({\text{10}}^{\text{-5}}\text{鈥�}m\right)}脳(\text{2.75}脳{\text{10}}^{\text{-10}}\text{鈥�}m)\\ =0.14\text{鈥�}mm\end{array}$

Hence, the distance of the first minimum from the center is$\text{0.14mm}$

(b)

To determine how many electrons per second are at the center of the detector, we solve using constructive interference.

For slit 1,

$\begin{array}{l}{\left|{\text{蠄}}_{\text{1}}\right|}^{\text{2}}鈭�\text{100}\\ \left|{\text{蠄}}_{\text{1}}\right|鈭�\text{10}\end{array}$鈥︹赌︹赌︹赌︹赌�..(4)

For slit 2,

$\begin{array}{l}{\left|{\text{蠄}}_2\right|}^{\text{2}}鈭�9\text{00}\\ \left|{\text{蠄}}_2\right|鈭�3\text{0}\end{array}$鈥︹赌︹赌︹赌︹赌︹赌︹赌︹赌�.(5)

Therefore, combining equations (4) and (5) for the constructive interference, we get the output of

$\begin{array}{c}\left|{\text{蠄}}_{\text{T}}\right|鈭�\left|{\text{蠄}}_1\right|+\left|{\text{蠄}}_2\right|\\ 鈭�\text{10}+\text{30}\\ 鈭�\text{40}\end{array}$

Therefore,

${\left|{\text{蠄}}_T\right|}^{\text{2}}鈭�16\text{00}$

(c)

To determine how many electrons per second are at detector X, we use equations (4) and (5) for the destructive interference, such that,

$\begin{array}{c}\left|{\text{蠄}}_{\text{D}}\right|鈭�\left|{\text{蠄}}_2\right|鈭�\left|{\text{蠄}}_1\right|\\ 鈭�\text{30}鈭�1\text{0}\\ 鈭�\text{20}\end{array}$

Therefore,

${\left|{\text{蠄}}_{\text{D}}\right|}^2鈭�\text{400}$