Physics · Dual Nature Of Radiation And Matter · NEET
Start from lambda = h/p. When an electron of charge e is accelerated from rest through V volts, the work done eV becomes its kinetic energy: KE = eV. Also KE = p squared / (2m), so p = root(2m e V). Put this into lambda = h/p to get lambda = h / root(2 m e V). Now plug in h = 6.63e-34, m = 9.1e-31, e = 1.6e-19. All the constants collapse into one number, giving lambda = 12.27/root(V) angstrom. So the whole formula is just lambda = h/p with the numbers filled in.
12.27/root(V) gives the answer in ANGSTROM (1 angstrom = 1e-10 m). If you want nanometres, use 1.227/root(V) nm, because 1 nm = 10 angstrom. Example: for V = 100 V, lambda = 12.27/10 = 1.227 angstrom = 0.1227 nm. Pick one form and stick to it so you do not lose a factor of 10.
No. For normal NEET voltages (a few volts up to a few kilovolts) the electron is non-relativistic, so KE = p squared/2m is exact enough. You never need the electron's speed or velocity separately. You only need V. Relativistic correction matters only above roughly 50 kV, which NEET does not ask.
lambda is proportional to 1/root(V). So if you make V four times bigger, root(V) doubles, and lambda becomes half. Higher accelerating voltage means a faster electron, more momentum, and a SHORTER de Broglie wavelength. This inverse-root link is a favourite NEET trap.
They are the same physics. Here the kinetic energy is supplied by the field, so KE = eV and lambda = h/root(2m e V). If a question just gives you the kinetic energy K directly (in joules or eV), use lambda = h/root(2mK). The accelerated-electron formula is the special case where K = eV. See the linked page on de Broglie wavelength in terms of energy.
An electron is accelerated through a potential difference of 10000 V. Its de Broglie wavelength is nearly (mass of electron = 9 x 10^-31 kg):
An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227 x 10^-2 nm, the potential difference is:
The de Broglie wavelength associated with an electron accelerated by a potential difference of 81 V is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
lambda = 12.27/root(100) = 12.27/10 = 1.227 angstrom, which equals 0.1227 nm. This is a very common check-value worth remembering.
lambda = h / root(2 m e V), where h is Planck's constant, m and e are the electron mass and charge, and V is the accelerating voltage. Numerically this becomes lambda = 12.27/root(V) angstrom.
The general form lambda = h/root(2 m q V) works for any charged particle. But the shortcut 12.27/root(V) angstrom is only for the electron, because it uses the electron mass and charge. A proton is much heavier, so its wavelength for the same V is far smaller.
More voltage means more kinetic energy and more momentum p. Since lambda = h/p, larger p gives smaller lambda. In fact lambda is proportional to 1/root(V).
Yes. For an electron, a kinetic energy of K electron-volts is the same as being accelerated through K volts, so lambda = 12.27/root(K) angstrom with K in eV. This is why the voltage and energy forms match.
