An electron accelerated through a potential difference V 1 has a de-Broglie wavelength of λ . When the potential is changed to V 2 , its de-Broglie wavelength increases by 50 % . The value of V 1 V 2 is equal to :

Step-by-step solution

Let initial wavelength be λ , then after 50 % increase wavelength will become 1 . 5 λ . Now, K E = P 2 2 m = e V   &   P = h λ Therefore, e V = h λ 2 2 m . So we can write e V 1 = h λ 2 2 m       . . . E q ( 1 ) and e V 2 = h 1 . 5 λ 2 2 m       . . . E q ( 2 ) From both equations, we get V 1 V 2 = 1 . 5 2 = 9 4