Let's solve each part step by step: --- ### (a) **Triangular Barrier and Electric Field** #### **Potential Plot and Expression** Given: \( U(x) = U_0 \left(1 - \frac{x}{d}\right)\Theta(x) \) Where: - \( U_0 \) = barrier height at \( x = 0 \) - \( \Theta(x) \) = unit step function \[ \Theta(x) = \begin{cases} 1, & x > 0 \\ 0, & x < 0 \end{cases} \] - For \( x < 0 \): \( U(x) = 0 \) - For \( 0 \leq x \leq d \): \( U(x) = U_0 \left(1 - \frac{x}{d}\right) \) (linearly decreases) - For \( x > d \): \( U(x) = 0 \) **Schematic Plot:** - At \( x=0 \), \( U = U_0 \). - At \( x = d \), \( U = 0 \). - For \( x < 0 \) or \( x > d \), \( U = 0 \). - So, the potential is a triangle starting at \( U_0 \) at \( x=0 \) and going to zero at \( x=d \). #### **Electric Field** \[ \mathcal{E}(x) = -\frac{\partial U(x)}{\partial x} \] For \( 0 < x < d \): \[ U(x) = U_0 \left(1 - \frac{x}{d}\right) \implies \frac{\partial U}{\partial x} = -\frac{U_0}{d} \] So, \[ \mathcal{E}(x) = \begin{cases} 0, & x < 0 \text{ or } x > d \\ \frac{U_0}{d}, & 0 < x < d \end{cases} \] --- ### (b) **Turning Points** Turning points: Where \( U(x) = E \). \[ U(x) = U_0 \left(1 - \frac{x}{d}\right) = E \] Solve for \( x \): \[ 1 - \frac{x}{d} = \frac{E}{U_0} \implies \frac{x}{d} = 1 - \frac{E}{U_0} \implies x = d\left(1 - \frac{E}{U_0}\right) \] - For \( x < 0 \), \( U(x) = 0 < E \) (not a turning point for positive \( E \)). - For \( x > d \), \( U(x) = 0 < E \). - So, the only relevant turning point for \( E < U_0 \) is at \( x_1=0 \) and \( x_2 = d(1-E/U_0) \). --- ### (c) **Calculate \( T(E) \):** \[ T(E) \approx \exp\left[-2 \int_{x_1}^{x_2} \kappa(x) dx \right] \] \[ \kappa(x) = \sqrt{\frac{2m[U(x) - E]}{\hbar^2}} \] For \( x_1 = 0 \), \( x_2 = d(1-E/U_0) \): \[ U(x) - E = U_0 \left(1 - \frac{x}{d}\right) - E = U_0 - U_0 \frac{x}{d} - E = (U_0 - E) - U_0\frac{x}{d} \] So, \[ \kappa(x) = \sqrt{\frac{2m}{\hbar^2} \left[(U_0 - E) - U_0 \frac{x}{d}\right]} \] Now, integrate: Let \( a = U_0 - E \), \( b = U_0/d \) \[ T(E) \approx \exp\left[-2 \int_0^{x_2} \sqrt{\frac{2m}{\hbar^2}(a - bx)} dx \right] \] Let \( y = a - bx \implies dx = -dy/b \). When \( x = 0, y = a \), when \( x = x_2, y = E \). \[ T(E) \approx \exp\left[-2 \int_{y=a}^{y=E} \sqrt{\frac{2m}{\hbar^2} y} \left(-\frac{1}{b}\right) dy \right] \] \[ = \exp\left[\frac{2}{b} \sqrt{\frac{2m}{\hbar^2}} \int_{E}^{a} \sqrt{y} dy \right] \] \[ = \exp\left[\frac{2}{b} \sqrt{\frac{2m}{\hbar^2}} \cdot \frac{2}{3} (a^{3/2} - E^{3/2}) \right] \] But \( a = U_0 - E \), \( b = U_0/d \): \[ T(E) = \exp\left[ \frac{4d}{3U_0} \sqrt{\frac{2m}{\hbar^2}} \left((U_0 - E)^{3/2} - E^{3/2}\right) \right] \] But for \( E \ll U_0 \), ignore the \( E^{3/2} \) term: \[ T(E) \approx \exp\left[ -\frac{4d}{3U_0} \sqrt{\frac{2m}{\hbar^2}} (U_0 - E)^{3/2} \right] \] --- ### (d) **Plot \( T(E) \) vs \( E \) Schematic** - \( T(E) \) is exponentially **small** for \( E \ll U_0 \). - \( T(E) \) rises **rapidly** as \( E \to U_0 \). - At \( E = U_0 \), \( T(E) \to 1 \). - The plot is a sharply rising exponential curve. --- ### (e) **Add Constant Barrier** The potential is: \[ U(x) = \begin{cases} 0, & x < 0 \\ U_0, & 0 < x < L \\ U_0\left(1 - \frac{x-L}{d}\right), & L < x < L+d \\ 0, & x > L+d \end{cases} \] So, you have a flat barrier from \( 0 \) to \( L \), then a triangular barrier from \( L \) to \( L + d \). **Schematic Plot**: - \( 0 \leq x < L \): Flat at \( U_0 \). - \( L \leq x \leq L + d \): Decreases linearly from \( U_0 \) to \( 0 \). - Elsewhere: \( U = 0 \). **\(\kappa(x)\) Integral**: \[ \int_{x_1}^{x_2} \kappa(x) dx = \int_{0}^{L} \sqrt{\frac{2m}{\hbar^2}(U_0 - E)} dx + \int_{L}^{L+d} \sqrt{\frac{2m}{\hbar^2}(U_0(1-\frac{x-L}{d}) - E)} dx \] First term: \[ \sqrt{\frac{2m}{\hbar^2}(U_0 - E)} \times L \] Second term: As before, but over \( x = L \) to \( L + d \): Let \( y = x - L \): \[ \int_{0}^{d} \sqrt{\frac{2m}{\hbar^2}(U_0(1 - \frac{y}{d}) - E)} dy \] Proceed as before. --- ### (f) **Critical Electric Field for Abrupt Transmission Change** Set \( E = U_0 / 2 \). We want the electric field \( \mathcal{E} \) such that the barrier is suppressed, i.e., the field term dominates the "flat" barrier (from part e). The triangular part becomes "thin" enough that transmission jumps. If \( \mathcal{E} = U_0/d \), the entire barrier can be suppressed for electrons with \( E = U_0/2 \). This is the operational principle in flash memories: - **Apply field:** Barrier lowers, electrons can tunnel (write). - **Remove field:** Barrier increases, electrons are stuck (data stored). --- ## **Summary of Answers** **(a)** - **Plot:** Triangle from \( U_0 \) at \( x=0 \) to 0 at \( x=d \). - **Field:** \( \mathcal{E}(x) = U_0/d \) for \( 0 < x < d \), zero elsewhere. **(b)** - **Turning points:** \( x_1 = 0 \), \( x_2 = d(1 - E/U_0) \). **(c)** \[ T(E) \approx \exp\left[ -\frac{4d}{3U_0} \sqrt{\frac{2m}{\hbar^2}} (U_0 - E)^{3/2} \right] \] **(d)** - **Plot:** Exponentially small for \( E \ll U_0 \), rises sharply as \( E \to U_0 \). **(e)** - **Plot:** Flat barrier (\( U_0 \)), then triangle. - **Integral:** Add \( L \sqrt{\frac{2m}{\hbar^2}(U_0 - E)} \) to earlier result. **(f)** - For \( E = U_0/2 \), need \( \mathcal{E} \approx U_0/d \) for abrupt transmission increase. This is the operational principle for flash memory. --- **If you need specific plots or more details on one step, let me know!**