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: Apply de Broglie formula.
\(\lambda = rac{h}{mv}\), with \(h = 6.6 imes10^{-34}\) J s, \(m = 9 imes10^{-31}\) kg, \(v = x imes10^6\) m/s.

Step 2: Calculate and convert to nm.
\(\lambda = rac{6.6 imes10^{-34}}{9 imes10^{-31} imes x imes10^6} = rac{6.6}{9x} imes10^{-9}\) m \(= rac{0.733}{x}\) nm \(\approx rac{0.73}{x}\) nm. \[ oxed{\dfrac{0.73}{x}} \]