Question #43227

An electron microscope has an objective of focal length $f_o$, an eyepiece of focal length $f_e$, and a tube length $L$. It is used to observe a biological sample of size $d_0 = \frac{d_{eye} f_o f_e}{L D}$, where $D$ is the least distance of distinct vision and $d_{eye}$ is the resolution limit of the human eye. The microscope can resolve the sample if the de Broglie wavelength of its accelerated electrons satisfies $\lambda_e \le d_0$.

The accelerating voltage $V(t)$ of the microscope is dynamically controlled by the instantaneous activity $A_Y(t)$ of isotope $Y$ in a successive radioactive decay series $X \xrightarrow{\lambda_X} Y \xrightarrow{\lambda_Y} Z$ (stable). At $t=0$, a pure sample of $N_0$ nuclei of $X$ is present. The decay constants are $\lambda_X = \lambda$ and $\lambda_Y = 2\lambda$.

The voltage is given by $V(t) = \alpha A_Y(t)$, where $\alpha$ is a calibration constant such that $\alpha \lambda N_0 = \frac{9 h^2}{8 m_e e d_0^2}$ (where $h$ is Planck's constant, $m_e$ is the electron mass, and $e$ is the elementary charge). Calculate the total time interval $\Delta t$ during which the microscope is capable of resolving the sample.