Doubly ionized lithium , , absorbs a photon and jumps from the ground state to its n=2 level. What was the wavelength of the photon?
When, doubly ionized lithium, , absorbs a photon and jumps from the ground state to its n=2 level, the wavelength of the photon 13.5 nm.
Whenever an electron absorbs energy enough for it to transition to higher state, it jumps to the higher state, and it loses energy while jumping down to the lower state from the higher state.
As you know that, energy of an electron in its nth orbit is given by
Where, z is the atomic number of hydrogen-like atom and n is the principal quantum number.
Hence, if the ion jumps from ground state to n=2 , the energy will be,
Energy of the photon is given by,
Here, h is the Planck’s constant, c is the speed of light, and is the wavelength.
The numerical value of hc is,
hc = 1240 eV .nm
Substitute known values into equation (1), and you have
It is shown in section 6.1 that for the E<U0 potential step, . Use it to calculate the probability density to the left of the step:
If things really do have a dual wave-particle nature, then if the wave spreads, the probability of finding the particle should spread proportionally, independent of the degree of spreading, mass, speed, and even Planck’s constant. Imagine that a beam of particles of mass m and speed v, moving in the x direction, passes through a single slit of width w . Show that the angle at which the first diffraction minimum would be found ( , from physical optics) is proportional to the angle at which the particle would likely be deflected , and that the proportionality factor is a pure number, independent of m, v, w and h . (Assume small angles: ).
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