The de Broglie wavelength for an electron accelerated through the potential difference of V1 volt is 1. When the potential difference is changed to V2 volt, the associated de Broglie wavelength is increased by 50%. If (V1/V2 )= (9/ ), then the value of is .
Answer
4
Solution
= h 2meV ⇒ 1 V ⇒ V1V2= ( 2 1 )2
Wavelength increased by 50% ⇒ 2=1.5 1= 32 1
⇒ V1V2= ( 32 )2= 94
Comparing with 9 : ∴ =4
Answer: 4