Question

The velocity of the electron in Bohr's first orbit is \(x\times10^6\ \text{m s}^{-1}\). The de Broglie wavelength associated with it (in nm) is
\[ \left(m_e=9\times10^{-31}\ \text{kg},\ h=6.6\times10^{-34}\ \text{J s}\right) \]

• A. \(\dfrac{x}{1.43}\)
• B. \(\dfrac{x}{0.73}\)
• C. \(\dfrac{0.73}{x}\)
• D. \(\dfrac{0.073}{x}\)

Hint:

For an electron moving with speed \(v\), \[ \lambda=\frac{h}{mv}. \] After substitution, convert metres to nanometres using \[ 1\ \text{nm}=10^{-9}\ \text{m}. \]

Answer 1

The Correct Option is C

Step 1: Use the de Broglie wavelength formula.
The de Broglie wavelength associated with a particle is given by \[ \lambda=\frac{h}{mv} \] where \[ h=6.6\times10^{-34}\ \text{J s} \] \[ m=9\times10^{-31}\ \text{kg} \] and \[ v=x\times10^6\ \text{m s}^{-1} \]

Step 2: Substitute the given values.
\[ \lambda = \frac{6.6\times10^{-34}} {(9\times10^{-31})(x\times10^6)} \] \[ = \frac{6.6\times10^{-34}} {9x\times10^{-25}} \] \[ = \frac{6.6}{9x}\times10^{-9}\ \text{m} \] \[ = \frac{0.733}{x}\times10^{-9}\ \text{m} \]

Step 3: Convert the wavelength into nanometres.
Since \[ 1\ \text{nm}=10^{-9}\ \text{m}, \] we get \[ \lambda = \frac{0.733}{x}\ \text{nm} \] \[ \lambda \approx \frac{0.73}{x}\ \text{nm} \]

Step 4: Final conclusion.
Therefore, the de Broglie wavelength associated with the electron is \[ \boxed{\frac{0.73}{x}\ \text{nm}} \]