The de Broglie wavelength of an electron in the 4^th orbit is:
(where, \( a_0\) = radius of 1^st orbit )
1. \(2 \pi a_0\)
2. \(8 \pi a_0\)
3. \(6 \pi a_0\)
4. \(4 \pi a_0\)

A proton and a Li³⁺ nucleus are accelerated by the same potential. If $\lambda$_Li³⁺ and ${\lambda }_p$ denote the de-Broglie wavelengths of Li³⁺ and proton respectively, then the value of \(\frac{\lambda _{Li^{3+}}}{\lambda _{p}}\) is x × 10⁻¹. The value of x is:

(Rounded off to the nearest integer)

(Mass of Li³⁺ = 8.3 the mass of a proton) 

1. 4 
2. 6 
3. 2
4. 8
 

Light of wavelength 248 nm is incident on a metal surface having a threshold energy of 3.0 eV.
Calculate the de Broglie wavelength (in Å) of the emitted photoelectrons. Round off your answer to the nearest integer.

Given:

√3 = 1.73

Planck's constant, h = 6.63 × 10⁻³⁴ J s

Mass of electron, mₑ = 9.1 × 10⁻³¹ kg

Speed of light, c = 3.0 × 10⁸ m s⁻¹

1 eV = 1.6 × 10⁻¹⁹ J

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 $h/\lambda$ ( where $\lambda$ is the wavelength associated with electron wave) is given by : 

1.  meV 

2.  2meV 

3. $\sqrt{meV}$

4. $\sqrt{2meV}$

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^–34 Js) is-:

1. 0.032 nm

2. 0.40 nm

3. 2.50 nm

4. 14.0 nm