Incorrect statement among the following is: 

The mass of a photon with wavelength 3.6 Å is :

Identify the matter waves with the shortest wavelength when traveling at the same speeds :

The wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m s^–1 is:

1. 662.6 × 10^–34 m

2. 66.26 × 10^–34 m

3. 6.626 × 10^–34 m

4. 6.626 × 10^–36 m

Kinetic Energy of an electron is 3.0×10^–25 J. The wavelength of the electron will be :

1. 886.7 nm

2. 896.7 nm

3. 990.7 nm

4. 880.7 nm

The kinetic energy of an electron is \(3.0 \times 10^{-25}~ \mathrm J.\) Its wave length would be: 
1. \(8.96 \times 10^{-7}~ \mathrm m\)
2. \(4.37 \times 10^{-6}~ \mathrm m\)
3. \(1.32 \times 10^{-7}~ \mathrm m\)
4. \(2.89 \times 10^{-4}~ \mathrm m\)

The wavelength of an electron moving with a velocity of 2.05 × 10⁷ m s^–1  would be:  

\(1 .\) \(4 . 65\) \(\times\) \(10^{- 12}\) \(m\)

\(2 .\) \(3 . 55\) \(\times\) \(10^{-11}\) \(m\)

\(3 .\) \(2 . 34\) \(\times\) \(10^{11}\) \(m\)

\(4 .\) \(6 . 43\) \(\times\) \(10^{ -11}\) \(m\)

The circumference of the Bohr orbit for the H atom is related to the de Broglie wavelength associated with the electron revolving around the orbit by the following relation: 
1. $2\mathrm{\pi r}=\mathrm{n\lambda }$
2. $\mathrm{\pi r}$ $=$ $2\mathrm{n\lambda }$
3. $\mathrm{mvr}$ $=$ $\mathrm{n\lambda }$
4. $\mathrm{vr}$ $=$ $2\mathrm{n\lambda }$

If the velocity of the electron is 1.6 × 10⁶ ${\mathrm{ms}}^{-1}$. The de Broglie wavelength associated with this electron is:  

$1.$ $590$ $\mathrm{pm}$
$2.$ $455$ $\mathrm{pm}$
$3.$ $512$ $\mathrm{pm}$
$4.$ $401$ $\mathrm{pm}$