given that tube is operated for t = 25 ms
which means the time for which it is used is t = 0.025 s
now the photons generated is given as
![N = 2.5 * 10^(14)](https://img.qammunity.org/2019/formulas/physics/high-school/thqwn3f8fzooj630guubmt73m24ryj0bdp.png)
each photon will produce 1 electron as we can assume it to be 100% efficient
so number of electrons per second will be same as number of photons per second
now in order to find the current we can say
current = rate of flow of charge
![i = (dN)/(dt)*e](https://img.qammunity.org/2019/formulas/physics/high-school/8xky0x0gtyaxkqazf1zrxzzezxykwcbus4.png)
![i = (2.5 * 10^(14))/(0.025)* 1.6 * 10^(-19)](https://img.qammunity.org/2019/formulas/physics/high-school/xjkbf6fz2blh7hagcih4h5bbib0eg93m3w.png)
![i = 1.6 * 10^(-3) A](https://img.qammunity.org/2019/formulas/physics/high-school/gbfrzkg8ctnc8q6niz3dg88uu0j1fzzklq.png)
so the current in the tube for given time will be 1.6 mA