Chemistry · Structure Of Atom · NEET
They are the same equation. If the question gives frequency (nu, in Hz), use E = h x nu. If it gives wavelength (lambda), use E = hc / lambda. This works because c = nu x lambda, so nu = c/lambda. Just look at what the question hands you: Hz means E = h nu, metres/nm means E = hc/lambda.
Multiply by 10^-9. So 400 nm = 400 x 10^-9 m = 4 x 10^-7 m. Always change lambda to metres first, or your answer will be wrong by a huge power of 10. Common ones: 1 nm = 10^-9 m, 1 pm = 10^-12 m, 1 Angstrom = 10^-10 m.
Look at E = hc/lambda. Lambda is in the bottom (denominator). When you divide by a smaller number, the answer gets bigger. So small lambda gives big E. That is why gamma rays and UV (tiny lambda) are dangerous, while radio waves (huge lambda) carry almost no energy per photon.
First find the energy of one photon using E = hc/lambda. Then multiply by Avogadro's number, 6.022 x 10^23. So E(mole) = (hc/lambda) x 6.022 x 10^23. NCERT does exactly this: energy of one photon x N_A gives energy per mole in joules.
With h in J s, c in m/s, and lambda in m, the energy comes out in joules (J). Write every number in scientific notation, handle the powers of 10 separately from the decimal parts, then combine. Example: 6.626e-34 x 3e8 = 1.988e-25; divide by 400e-9 = 4e-7 gives about 4.97e-19 J.
Power (watt) is energy per second. First find how much light energy the bulb gives per second. Then divide that by the energy of one photon (E = hc/lambda) to get the number of photons per second. Number of photons = total light energy / energy of one photon.
A bulb is rated at 150 watt, converting 8% of its energy into light. If the energy of one photon is 4.42 x 10^-19 J, how many photons are emitted by the bulb per second?
A station of All India Radio broadcasts on a frequency of 1368 kHz. The wavelength of the electromagnetic radiation emitted is: (c = 3 x 10^8 m/s)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
E = h x nu = hc / lambda, where h = 6.626 x 10^-34 J s, c = 3 x 10^8 m/s, nu is frequency in Hz, and lambda is wavelength in metres. It gives the energy of one single photon in joules.
Inversely proportional. E = hc/lambda, so as lambda increases E decreases. Photon energy is directly proportional to frequency (E = h nu).
h x c = 6.626 x 10^-34 J s x 3 x 10^8 m/s = 1.988 x 10^-25 J m (often rounded to about 1.99 x 10^-25 J m). Dividing this by lambda in metres gives energy in joules.
In Einstein's equation h nu = h nu0 + (1/2)m v^2, the h nu term is the incoming photon energy from E = hc/lambda. If this energy beats the work function, an electron is ejected. So this same formula is the base for photoelectric NEET numericals.
It combines several ideas in one sum: E = hc/lambda, unit conversion (nm to m), bulb power, and moles of photons. One clean formula can be asked many ways, so mastering E = hc/lambda covers a whole cluster of NEET questions.
