The work of a metal is W-0 and lambda is the wavelength of the incidence light- There will be no photoelectric effect -

Question

The work of a metal is W-0 and lambda  is the wavelength of the incidence light- There will be no photoelectric effect -

Toppr Admin · Accepted Answer

Correct option is A- -lambda -frac-hc-W-0-begin-array-l-k-max -h v-phi -text - Since there is no photo-electric effect - -therefore h v-phi-lt-0 -frac-h c-lambda-b-w-0-lt-0 -W-0-gt-frac-h c-lambda-Rightarrow-left-lambda-gt-frac-h c-omega-0-right-end-array-

The work function of a metal is $$W_0$$ and $$\lambda $$ is the wavelength of the incidence light. There will be no photoelectric effect if :

Correct option is A. $$\lambda >\frac{hc}{W_0}$$
$$\begin{array}{l}k_{\max }=h v-\phi \\\text { Since there is no photo-electric effect } \\\therefore h v\phi<0 \\\frac{h c}{\lambda_{b}}-w_{0}<0 \\W_{0}>\frac{h c}{\lambda}\Rightarrow\left[\lambda>\frac{h c}{\omega_{0}}\right]\end{array}$$

The work function of a metal is $$W_0$$ and $$\lambda $$ is the wavelength of the incidence light. There will be no photoelectric effect if :

Correct option is A. $$\lambda >\frac{hc}{W_0}$$$$\begin{array}{l}k_{\max }=h v-\phi \\\text { Since there is no photo-electric effect } \\\therefore h v\phi<0 \\\frac{h c}{\lambda_{b}}-w_{0}<0 \\W_{0}>\frac{h c}{\lambda}\Rightarrow\left[\lambda>\frac{h c}{\omega_{0}}\right]\end{array}$$

Correct option is A. $$\lambda >\frac{hc}{W_0}$$
$$\begin{array}{l}k_{\max }=h v-\phi \\\text { Since there is no photo-electric effect } \\\therefore h v\phi<0 \\\frac{h c}{\lambda_{b}}-w_{0}<0 \\W_{0}>\frac{h c}{\lambda}\Rightarrow\left[\lambda>\frac{h c}{\omega_{0}}\right]\end{array}$$