Skip to content
dronacharya.ai
← JEE Main 2022 · 30 Jul · Shift 2
PhysicsSingle correct+4 / -1Question 19 of 90

JEE Main 2022 · 30 Jul · Shift 2, Physics question 19

The de Broglie wavelength associated with an electron, accelerated through a potential difference of 15056 V, is:

Choose one option
Answer and worked solution

Official answer: A

Electron λ=12.27V\lambda=\dfrac{12.27}{\sqrt V} Å =12.27122.7=0.1=\dfrac{12.27}{122.7}=0.1 Å.

(A)

Topics:de Broglie waves
← Previous questionNext question →