JEE PYQ: Experimental Physics Question 24
Question 24 - 2019 (10 Jan 2019 Shift 1)
In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5 \times 10^{-12}$ m, the minimum electron energy required is close to:
(1) 500 keV
(2) 100 keV
(3) 1 keV
(4) 25 keV
Show Answer
Answer: (4)
Solution
$\lambda = \frac{h}{P}$. KE $= \frac{P^2}{2m} = \frac{(h/\lambda)^2}{2m} = \frac{(6.6 \times 10^{-34})^2}{(7.5 \times 10^{-12})^2 \times 2 \times 9.1 \times 10^{-31}} = 25$ keV.
BETA
