# Poisson parameter in Merton's Jump-Diffusion Model to price call option

I've been taught the following European call valuation formula under jump-diffusion model: $$\begin{equation} price = E[e^{-rT}max(S_T-K,0)] =\sum_{j = 0}^\infty e^{-rT}P_j(\lambda)E[max(S_T-K,0)|J=j] \end{equation}$$ where $$J$$ is the number of jumps, probability term $$P$$ is the poisson process: $$\begin{equation} e^{-\lambda T} \frac{(\lambda T)^j}{j!} \end{equation}$$ And I've been given directly the final form of the valuation formula: $$\begin{equation} price = \sum_{j = 0}^\infty P_j(\lambda^{'})BS(S_0,r_j,\sigma_j^2) \end{equation}$$ where: $$\begin{equation} \lambda^{'}=\lambda(1+k) \end{equation}$$ $$k$$ is the average jump size, and: $$\begin{equation} r_j = r-\lambda k+j\ln(1+k) / T, \sigma_j^2=\sigma^2+j\sigma_s^2/T \end{equation}$$ $$\sigma^2$$ is original Black-Scholes volitility, $$\sigma_s^2$$ is the jump variance.

By expanding the Black-Scholes equation, I've managed to derive $$r_j$$ and $$\sigma_j^2$$, but I'm confused about where $$\lambda^{'}$$ comes from. Can someone help derive this term?

In case of anything unclear, the formula can also be found here and here (in the second link, $$m=1+k$$)