The equation λ=1.227/x nm can be used to find the de-Brogli wavelength of an electron. In this equation x stands for :
Where
m= mass of electron
P= momentum of electron
K= Kinetic energy of electron
V= Accelerating potential in volts for electron

Solution

Physics sir · 3 steps

That 1.227 is just h, m and e worked out for an electron, so the only variable left inside the root is V.

1. Energy from voltage. An electron accelerated from rest through a potential V gains kinetic energy K = eV. Since K = p²/2m, its momentum is p = √2meV.

   λ = h/p = h/√2meV = h/√2me × 1/√V

2. Work out constant. With h = 6.626 × 10⁻³⁴ J s, m = 9.11 × 10⁻³¹ kg and e = 1.602 × 10⁻¹⁹ C, you get √2me = 5.40 × 10⁻²⁵. So h/√2me = 1.227 × 10⁻⁹ m = 1.227 nm.
3. So. λ = 1.227/√V nm with V in volts, so x = √V, option (d).

The idea to keep. For an electron accelerated through V volts, remember λ = 1.227/√V nm. The fixed constants h, m and e are already packed into the 1.227.

Answer from the NTA final answer key. Solution by PrepWiser.