NEET + JEEAdvanced
In a singly-ionized helium atom (He+, atomic number Z = 2) the…
Question
In a singly-ionized helium atom (He+, atomic number Z = 2) the electron makes a transition from the n = 3 level to the n = 1 level. Using ground-state hydrogen energy 13.6 eV, find the energy (in eV) of the emitted photon.
✓ Verified answer: 48.3556checked by our engine — not a guess
Step-by-step solution
Energy levels: E_n = -13.6 Z^2 / n^2 eV. For He+, Z^2 = 4.
Photon energy = 13.6*4*(1/1^2 - 1/3^2) = 54.4*(1 - 1/9) = 54.4*(8/9) = 48.3556 eV.
Final answer48.3556
Stuck on a problem like this?
Paste any JEE or NEET question — verified working, a confidence %, and an honest “not sure” instead of a bluff.
Solve my doubt →More Neet Modern Thermo solutions
Light of a single wavelength falls on two different metal surfaces.…NEET + JEE · PhysicsA hypothetical photoelectric experiment uses the same monochromatic…NEET + JEE · PhysicsFor the n = 2 stationary orbit of a hydrogen atom (Bohr radius a0 =…NEET + JEE · PhysicsA radioactive sample has an initial activity of 8000 disintegrations…NEET + JEE · PhysicsTwo deuterium nuclei fuse to form a helium-4 nucleus. Given atomic…NEET + JEE · PhysicsA Zener diode of breakdown voltage 6 V is used as a voltage…NEET + JEE · PhysicsAn n-p-n transistor in common-emitter configuration has a current…NEET + JEE · PhysicsAt what temperature (in kelvin) would the rms speed of oxygen…NEET + JEE · Physics