Question

The photoelectric effect of Ba consists of emission of electrons from its ionisation energy on irradiation with light from the ionization energy of $He$-atom is irradiation with light. A photon with a minimum energy of $3.97\times {10}^{-19}J$ is necessary to eject electron from $Ba$.
(a)What is the frequency of radiation corresponding to this value?
(b) Will a blue light of wavelength $450nm$ be able to eject at electron from $Ba$? (Given $h=6.626\times {10}^{-27}erg-sec$)

• A. $6\times {10}^{13}Hz$, yes
• B. $6\times {10}^{13}Hz$, no
• C. $5.99\times {10}^{14}Hz$, yes
• D. $3.6\times {10}^{13}Hz$, no

Answer 1

The correct option is C $5.99\times {10}^{14}Hz$, yes

(a) The frequency of radiation corresponding to the energy $3.97\times {10}^{-19}J$ is given by -
$\upsilon =E/h$ where E = energy , h= planck's constant
Also we know that,
$1J={10}^{7}erg$
so, $3.97\times {10}^{-19}J=3.97\times {10}^{-12}erg$
Therefore, $\upsilon =\frac{3.97\times {10}^{-12}erg}{6.626\times {10}^{-27}erg-sec}$
$\upsilon =$5.99\times 10^{14} Hz
(b) Energy corresponding to the wavelength 450 nm :
We know $E=hc/\lambda$
where E = energy , h= planck's constant , $\lambda$ = wavelength
$E=\frac{6.626\times {10}^{-27}erg-sec\times 3\times {10}^{8}m}{450\times {10}^{-9}m}$
$E=4.41\times {10}^{-12}erg$
Since this energy is greater than the given threshold energy $3.97\times {10}^{-12}erg$, thus the blue light of the given wavelength will be able to eject the electrons.