A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm, the other is at

• 1035 nm

• 325 nm

• 743 nm 

• 518 nm

A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/λ (where λ is wavelength associated with electron wave) is given by:

• 2meV

• 
• 

• mev

C.

The relation between h/λ and energy is given as:
Applying de-Broglie wavelength and kinetic energy term in eV.
de-Broglie wavelength for an electron (λ) = h/p
⇒ p  = h/ λ       (i)
Kinetic energy of an electron = eV
As we know that,

From equations (i) and (ii), we get

According to Bohr’s theory, the angular momentum of an electron in 5th orbit is

• 25h/ π

• 1.0h/π

• 10/π

• 2.5h/π

D.

2.5h/π

Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 × 10³ ms^–1
(Mass of proton = 1.67 × 10^–27 kg and h = 6.63 ×10^–34Js)

• 0.032 nm

• 0.40 nm

• 2.5 nm

• 14.0 nm

B.

0.40 nm

Consider the ground state of Cr atom (Z = 24). The number of electrons with the azimuthal quantum numbers I =1 and 2 are respectively

• 12 and 4

• 16 and 5

• 16 and 4

• 12 and 5