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Chemistry Question on Bohr’s Model for Hydrogen Atom

A light of frequency 1.6×1016 Hz1.6 \times 10^{16} \, Hz when falls on a metal plate emits electrons that have double the kinetic energy compared to the kinetic energy of emitted electrons when frequency of 1.0×1016 Hz1.0 \times 10^{16} \, Hz falls on the same plate. The threshold frequency (v0)(v_0) of the metal in HzHz is

A

1×10151 \times 10^{15}

B

4×10154 \times 10^{15}

C

3×10153 \times 10^{15}

D

4×10154 \times 10^{15}

Answer

4×10154 \times 10^{15}

Explanation

Solution

When 1.6×10161.6 \times 10^{16} Hz frequency falls on metal plate than double kinetic energy obtained.

h(1.6×1016−v0)=2K.E⟶h\left(1.6 \times 10^{16}-v_{0}\right)=2 K.E \longrightarrow Eq .(i)

When 1.0×1016 Hz1.0 \times 10^{16}\, Hz frequency falls on same plate than K.E.

h(1.0×1016−v0)=K.E⟶h\left(1.0 \times 10^{16}-v_{0}\right)= K.E \longrightarrow (Eq .(ii))

From Eqs. (i) and (ii)

h(1.6×1016−v0)h(1.0×1016−v0)=2K⋅EK.E\frac{h\left(1.6 \times 10^{16}-v_{0}\right)}{h\left(1.0 \times 10^{16}-v_{0}\right)}=\frac{2 K \cdot E }{ K . E }
1.6×1016−v0=2×1.0×1016−v01.6 \times 10^{16}-v_{0}=2 \times 1.0 \times 10^{16}-v_{0}
1.6×1016−2(1.0×1016)=4×1015 Hz1.6 \times 10^{16}-2\left(1.0 \times 10^{16}\right)=4 \times 10^{15}\,Hz
v0=4×1015 Hzv_{0}=4 \times 10^{15}\, Hz