Photons of frequency 3.2 × 1016 Hz is used to irradiate a metal surface, the maximum kinetic energy of the emitted photo-electron is (3 th /4). of the energy of the irradiating photon? What is the threshold frequency of the metal?

Question

Photons of frequency 3.2 × 1016 Hz is used to irradiate a metal surface, the maximum kinetic energy of the emitted photo-electron is (3 th /4). of the energy of the irradiating photon? What is the threshold frequency of the metal? (A) 2.4 × 1025 Hz (B) 2.4 × 1016 Hz (C) 1.6 × 1015 Hz (D) 8 × 1015 Hz

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/f6f52f2d626b8cdd7e33e61b185d3b64-.png" /> Apply, h v-h v0=k E Apply, h v-h v0=(3/4) h v ⇒ h v-(3/4) h v=h v0 (h v/4)=h v0 ⇒ (v/4)=v0 (3.2 × 1016/4)=v0 =8 × 1015 Hz

Q. Photons of frequency $3.2 \times 1 0^{16} Hz$ is used to irradiate a metal surface, the maximum kinetic energy of the emitted photo-electron is $\frac{3 ^{t h}}{4}$ . of the energy of the irradiating photon? What is the threshold frequency of the metal?

$2.4 \times 1 0^{25} Hz$

$2.4 \times 1 0^{16} Hz$

$1.6 \times 1 0^{15} Hz$

$8 \times 1 0^{15} Hz$

Apply, $h v - h v_{0} = k E$

Apply, $h v - h v_{0} = \frac{3}{4} h v$

$\Rightarrow h v - \frac{3}{4} h v = h v_{0}$

$\frac{h v}{4} = h v_{0}$

$\Rightarrow \frac{v}{4} = v_{0}$

$\frac{3.2 \times 1 0 ^{16}}{4} = v_{0}$

$= 8 \times 1 0^{15} Hz$

Q. Photons of frequency 3.2×1016Hz is used to irradiate a metal surface, the maximum kinetic energy of the emitted photo-electron is 43th​. of the energy of the irradiating photon? What is the threshold frequency of the metal?

2.4×1025Hz

2.4×1016Hz

1.6×1015Hz

8×1015Hz